Class 11: Limits – Exercise 29.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Evaluate the following limits:

$\displaystyle \text{Question 1: } \lim \limits_{x \to 0} \frac{\sqrt{1+x+x^2} -1}{ x}$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{\sqrt{1+x+x^2} -1}{ x} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{1+x+x^2} -1}{ x} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(\sqrt{1+x+x^2} -1)(\sqrt{1+x+x^2} +1)}{ x(\sqrt{1+x+x^2} +1)}\Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{1+x+x^2 -1}{ x(\sqrt{1+x+x^2} +1)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{x(1+x)}{ x(\sqrt{1+x+x^2} +1)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(1+x)}{ (\sqrt{1+x+x^2} +1)} \Bigg)$

$\displaystyle = \frac{1+0}{\sqrt{1+0+0} +1}$

$\displaystyle = \frac{1}{2}$

$\\$

$\displaystyle \text{Question 2: } \lim \limits_{x \to 0} \frac{2x}{ \sqrt{a+x} - \sqrt{a-x}}$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{2x}{ \sqrt{a+x} - \sqrt{a-x}} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{2x}{ \sqrt{a+x} - \sqrt{a-x}} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{2x}{ \sqrt{a+x} - \sqrt{a-x}} \times \frac{\sqrt{a+x} + \sqrt{a-x}}{ \sqrt{a+x} + \sqrt{a-x}} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{2x(\sqrt{a+x} + \sqrt{a-x})}{ (a+x)-(a-x)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{2x(\sqrt{a+x} + \sqrt{a-x})}{ 2x} \Bigg)$

$\displaystyle = \sqrt{a} +\sqrt{a}$

$\displaystyle = 2\sqrt{a}$

$\\$

$\displaystyle \text{Question 3: } \lim \limits_{x \to 0} \frac{\sqrt{a^2+x^2}-a}{x^2}$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{\sqrt{a^2+x^2}-a}{x^2} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{a^2+x^2}-a}{x^2} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(\sqrt{a^2+x^2}-a)(\sqrt{a^2+x^2}+a)}{x^2(\sqrt{a^2+x^2}+a)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{a^2+x^2-a^2}{x^2(\sqrt{a^2+x^2}+a)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{1}{(\sqrt{a^2+x^2}+a)} \Bigg)$

$\displaystyle = \frac{1}{\sqrt{a^2}+1}$

$\displaystyle = \frac{1}{2a}$

$\\$

$\displaystyle \text{Question 4: } \lim \limits_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{2x }$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{\sqrt{1+x}-\sqrt{1-x}}{2x } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{1+x}-\sqrt{1-x}}{2x } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{2x(\sqrt{1+x}+\sqrt{1-x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(1+x)-(1-x)}{2x(\sqrt{1+x}+\sqrt{1-x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{2x}{2x(\sqrt{1+x}+\sqrt{1-x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{1}{(\sqrt{1+x}+\sqrt{1-x}) } \Bigg)$

$\displaystyle = \frac{1}{\sqrt{1+0} + \sqrt{1-0}}$

$\displaystyle = \frac{1}{2}$

$\\$

$\displaystyle \text{Question 5: } \lim \limits_{x \to 2} \frac{\sqrt{3-x} -1}{2- x}$

Answer:

$\displaystyle \text{When } x = 2, \text{ the expression } \frac{\sqrt{3-x} -1}{2- x} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 2} \Bigg( \frac{\sqrt{3-x} -1}{2- x} \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{(\sqrt{3-x} -1)(\sqrt{3-x} +1)}{(2- x)(\sqrt{3-x} +1)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{3-x-1}{(2- x)(\sqrt{3-x} +1)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{(2-x)}{(2- x)(\sqrt{3-x} +1)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{1}{(\sqrt{3-x} +1)} \Bigg)$

$\displaystyle = \frac{1}{\sqrt{3-2}+1}$

$\displaystyle = \frac{1}{2}$

$\\$

$\displaystyle \text{Question 6: } \lim \limits_{x \to 3} \frac{x-3}{ \sqrt{x-2} - \sqrt{4-x}}$

Answer:

$\displaystyle \text{When } x = 3, \text{ the expression } \frac{x-3}{ \sqrt{x-2} - \sqrt{4-x}} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 3} \Bigg( \frac{x-3}{ \sqrt{x-2} - \sqrt{4-x}} \Bigg)$

$\displaystyle = \lim \limits_{x \to 3} \Bigg( \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{ (\sqrt{x-2} + \sqrt{4-x})(\sqrt{x-2} - \sqrt{4-x})} \Bigg)$

$\displaystyle = \lim \limits_{x \to 3} \Bigg( \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{ (x-2)-(4-x)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 3} \Bigg( \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{ 2(x-3)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 3} \Bigg( \frac{(\sqrt{x-2} + \sqrt{4-x})}{ 2} \Bigg)$

$\displaystyle = \Bigg( \frac{\sqrt{3-2} + \sqrt{4-3}}{2} \Bigg)$

$\displaystyle = \Bigg( \frac{\sqrt{1} + \sqrt{1}}{2} \Bigg)$

$\displaystyle = 1$

$\\$

$\displaystyle \text{Question 7: } \lim \limits_{x \to 0} \frac{x}{ \sqrt{1+x} - \sqrt{1-x}}$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{x}{ \sqrt{1+x} - \sqrt{1-x}} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{x}{ \sqrt{1+x} - \sqrt{1-x}} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(x)(\sqrt{1+x} + \sqrt{1-x})}{ (\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(x)(\sqrt{1+x} + \sqrt{1-x})}{ (1+x)-(1-x)}\Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(x)(\sqrt{1+x} + \sqrt{1-x})}{ 2x} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{1+x} + \sqrt{1-x}}{ 2} \Bigg)$

$\displaystyle = \frac{\sqrt{1}+\sqrt{1}}{2}$

$\displaystyle = 1$

$\\$

$\displaystyle \text{Question 8: } \lim \limits_{x \to 1} \frac{\sqrt{5x-4}-\sqrt{x}}{x-1 }$

Answer:

$\displaystyle \text{When } x = 1, \text{ the expression } \frac{\sqrt{5x-4}-\sqrt{x}}{x-1 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{\sqrt{5x-4}-\sqrt{x}}{x-1 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(\sqrt{5x-4}-\sqrt{x})(\sqrt{5x-4}+\sqrt{x})}{(x-1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{5x-4-x}{(x-1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{4(x-1)}{(x-1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{4}{(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \frac{4}{\sqrt{5-4} + \sqrt{1}}$

$\displaystyle = \frac{4}{2}$

$\displaystyle = 2$

$\\$

$\displaystyle \text{Question 9: } \lim \limits_{x \to 1} \frac{x-1}{ \sqrt{x^2+3} - 2}$

Answer:

$\displaystyle \text{When } x = 1, \text{ the expression } \frac{x-1}{ \sqrt{x^2+3} - 2} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{(x-1)(\sqrt{x^2+3} + 2)}{ (\sqrt{x^2+3} - 2)(\sqrt{x^2+3} + 2)} \Bigg)$

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{(x-1)(\sqrt{x^2+3} + 2)}{ x^2+3-4} \Bigg)$

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{(x-1)(\sqrt{x^2+3} + 2)}{ (x-1)(x+1)} \Bigg)$

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{\sqrt{x^2+3} + 2}{ x+1} \Bigg)$

$\displaystyle = \frac{\sqrt{1+3} + 2}{1+1}$

$\displaystyle = \frac{4}{2}$

$\displaystyle = 1$

$\\$

$\displaystyle \text{Question10: } \lim \limits_{x \to 3} \frac{\sqrt{x+3}-\sqrt{6}}{x^2-9 }$

Answer:

$\displaystyle \text{When } x = 3, \text{ the expression } \frac{\sqrt{x+3}-\sqrt{6}}{x^2-9 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 3} \Bigg( \frac{\sqrt{x+3}-\sqrt{6}}{x^2-9 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 3} \Bigg( \frac{(\sqrt{x+3}-\sqrt{6})(\sqrt{x+3}+\sqrt{6})}{(x^2-9)(\sqrt{x+3}+\sqrt{6}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 3} \Bigg( \frac{x+3-6}{(x-3)(x+3)(\sqrt{x+3}+\sqrt{6}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 3} \Bigg( \frac{(x-3)}{(x-3)(x+3)(\sqrt{x+3}+\sqrt{6}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 3} \Bigg( \frac{1}{(x+3)(\sqrt{x+3}+\sqrt{6}) } \Bigg)$

$\displaystyle = \frac{1}{6(\sqrt{6}+\sqrt{6})}$

$\displaystyle = \frac{1}{12\sqrt{6}}$

$\\$

$\displaystyle \text{Question11: } \lim \limits_{x \to 1} \frac{\sqrt{5x-4}-\sqrt{x}}{x^2-1 }$

Answer:

$\displaystyle \text{When } x = 1, \text{ the expression } \frac{\sqrt{5x-4}-\sqrt{x}}{x^2-1 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{\sqrt{5x-4}-\sqrt{x}}{x^2-1 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(\sqrt{5x-4}-\sqrt{x})(\sqrt{5x-4}+\sqrt{x})}{(x^2-1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{5x-4-x}{(x-1)(x+1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{4(x-1)}{(x-1)(x+1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{4}{(x+1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \frac{4}{(\sqrt{5-4}+\sqrt{1})(1+1)}$

$\displaystyle = \frac{4}{2 \times 2}$

$\displaystyle = 1$

$\\$

$\displaystyle \text{Question 12: } \lim \limits_{x \to 0} \frac{\sqrt{1+x} -1}{ x}$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{\sqrt{1+x} -1}{ x} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{1+x} -1}{ x} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(\sqrt{1+x} -1)(\sqrt{1+x} +1)}{ x(\sqrt{1+x} +1)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{1+x-1}{ x(\sqrt{1+x} +1)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{1}{ (\sqrt{1+x} +1)} \Bigg)$

$\displaystyle = \frac{1}{\sqrt{1+0}+1}$

$\displaystyle = \frac{1}{2}$

$\\$

$\displaystyle \text{Question13: } \lim \limits_{x \to 2} \frac{\sqrt{x^2+1}-\sqrt{5}}{x-2 }$

Answer:

$\displaystyle \text{When } x = 2, \text{ the expression } \frac{\sqrt{x^2+1}-\sqrt{5}}{x-2 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 2} \Bigg( \frac{\sqrt{x^2+1}-\sqrt{5}}{x-2 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{(\sqrt{x^2+1}-\sqrt{5})(\sqrt{x^2+1}+\sqrt{5})}{(x-2)(\sqrt{x^2+1}+\sqrt{5}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{x^2+1-5}{(x-2)(\sqrt{x^2+1}+\sqrt{5}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{x^2-4}{(x-2)(\sqrt{x^2+1}+\sqrt{5}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{(x-2)(x+2)}{(x-2)(\sqrt{x^2+1}+\sqrt{5}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{x+2}{\sqrt{x^2+1}+\sqrt{5} } \Bigg)$

$\displaystyle = \frac{4}{2\sqrt{5}}$

$\displaystyle = \frac{2}{\sqrt{5}}$

$\\$

$\displaystyle \text{Question 14: } \lim \limits_{x \to 2} \frac{x-2}{ \sqrt{x} - \sqrt{2}}$

Answer:

$\displaystyle \text{When } x = 2, \text{ the expression } \frac{x-2}{ \sqrt{x} - \sqrt{2}} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 2} \Bigg( \frac{x-2}{ \sqrt{x} - \sqrt{2}} \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{(\sqrt{x})^2-(\sqrt{2})^2}{ \sqrt{x} - \sqrt{2}} \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{ (\sqrt{x} - \sqrt{2})} \Bigg)$

$\displaystyle = \sqrt{2}+\sqrt{2}$

$\displaystyle = 2\sqrt{2}$

$\\$

$\displaystyle \text{Question 15: } \lim \limits_{x \to 7} \frac{4-\sqrt{9+x}}{ 1 - \sqrt{8-x}}$

Answer:

$\displaystyle \text{When } x = 7, \text{ the expression } \frac{4-\sqrt{9+x}}{ 1 - \sqrt{8-x}} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 7} \Bigg( \frac{4-\sqrt{9+x}}{ 1 - \sqrt{8-x}} \Bigg)$

$\displaystyle = \lim \limits_{x \to 7} \Bigg( \frac{(4-\sqrt{9+x})(4+\sqrt{9+x})}{ (4+\sqrt{9+x})} \times \frac{ (1 + \sqrt{8-x})}{(1-\sqrt{8-x})((1 + \sqrt{8-x})} \Bigg)$

$\displaystyle = \lim \limits_{x \to 7} \Bigg( \frac{16-(9+x)}{ 4+\sqrt{9+x}} \times \frac{ 1 - \sqrt{8-x}}{1-(8-x)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 7} \Bigg( \frac{ -1(-7+x)(1 + \sqrt{8-x}) }{ (4+\sqrt{9+x})(-7+x) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 7} \Bigg( \frac{ -1(1 + \sqrt{8-x}) }{ (4+\sqrt{9+x}) } \Bigg)$

$\displaystyle = - \frac{1 + \sqrt{8-7} }{4 + \sqrt{9+7} }$

$\displaystyle = \frac{-2}{4+4}$

$\displaystyle = \frac{-1}{4}$

$\\$

$\displaystyle \text{Question 16: } \lim \limits_{x \to 0} \frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax} }$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax} } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax} } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(\sqrt{a+x}-\sqrt{a})(\sqrt{a+x}+\sqrt{a})}{(x\sqrt{a^2+ax}) (\sqrt{a+x}+\sqrt{a})} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{a+x-a}{(x\sqrt{a^2+ax}) (\sqrt{a+x}+\sqrt{a})} \Bigg)$

$\displaystyle = \frac{1}{(\sqrt{a^2})(\sqrt{a}+\sqrt{a})}$

$\displaystyle = \frac{1}{2a\sqrt{a}}$

$\\$

$\displaystyle \text{Question 17: } \lim \limits_{x \to 5} \frac{x-5}{ \sqrt{6x-5} - \sqrt{4x+5}}$

Answer:

$\displaystyle \text{When } x = 5, \text{ the expression } \frac{x-5}{ \sqrt{6x-5} - \sqrt{4x+5}} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 5} \Bigg( \frac{x-5}{ \sqrt{6x-5} - \sqrt{4x+5}} \Bigg)$

$\displaystyle = \lim \limits_{x \to 5} \Bigg( \frac{(x-5)(\sqrt{6x-5} + \sqrt{4x+5})}{ (\sqrt{6x-5} - \sqrt{4x+5})(\sqrt{6x-5} + \sqrt{4x+5})} \Bigg)$

$\displaystyle = \lim \limits_{x \to 5} \Bigg( \frac{(x-5)(\sqrt{6x-5} + \sqrt{4x+5})}{ (6x-5) - ( 4x+5)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 5} \Bigg( \frac{(x-5)(\sqrt{6x-5} + \sqrt{4x+5})}{ 2x-10} \Bigg)$

$\displaystyle = \lim \limits_{x \to 5} \Bigg( \frac{(x-5)(\sqrt{6x-5} + \sqrt{4x+5})}{ 2(x-5)} \Bigg)$

$\displaystyle = \frac{ \sqrt{6\times 5 - 5}+ \sqrt{4 \times 5 + 5}}{ 2}$

$\displaystyle = \frac{5+5}{2}$

$\displaystyle = 5$

$\\$

$\displaystyle \text{Question18: } \lim \limits_{x \to 1} \frac{\sqrt{5x-4}-\sqrt{x}}{x^3-1 }$

Answer:

$\displaystyle \text{When } x =1, \text{ the expression } \frac{\sqrt{5x-4}-\sqrt{x}}{x^3-1 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{\sqrt{5x-4}-\sqrt{x}}{x^3-1 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(\sqrt{5x-4}-\sqrt{x})(\sqrt{5x-4}+\sqrt{x})}{(x^3-1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{5x-4-x}{(x-1)(x^2+x+1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{4(x-1)}{(x-1)(x^2+x+1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{4}{(x^2+x+1)(\sqrt{5x-4}+\sqrt{x}) } \Bigg)$

$\displaystyle = \frac{4}{(1+1+1)(\sqrt{1}+\sqrt{1})}$

$\displaystyle = \frac{2}{3}$

$\\$

$\displaystyle \text{Question 19: } \lim \limits_{x \to 2} \frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2 }$

Answer:

$\displaystyle \text{When } x = 2, \text{ the expression } \frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 2} \Bigg( \frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{(\sqrt{1+4x}-\sqrt{5+2x})(\sqrt{1+4x}+\sqrt{5+2x})}{(x-2)(\sqrt{1+4x}+\sqrt{5+2x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{(1+4x)-(5+2x)}{(x-2)(\sqrt{1+4x}+\sqrt{5+2x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{2(x-2)}{(x-2)(\sqrt{1+4x}+\sqrt{5+2x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{2}{(\sqrt{1+4x}+\sqrt{5+2x}) } \Bigg)$

$\displaystyle = \frac{2}{\sqrt{1+4\times 2} + \sqrt{5+2\times2}}$

$\displaystyle = \frac{2}{3+3}$

$\displaystyle = \frac{1}{3}$

$\\$

$\displaystyle \text{Question 20: } \lim \limits_{x \to 1} \frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1 }$

Answer:

$\displaystyle \text{When } x = 1, \text{ the expression } \frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(\sqrt{3+x}-\sqrt{5-x})(\sqrt{3+x}\sqrt{5-x})}{(x^2-1)(\sqrt{3+x}+\sqrt{5-x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(3+x)-(5-x)}{(x-1)(x+1)(\sqrt{3+x}+\sqrt{5-x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{2(x-1)}{(x-1)(x+1)(\sqrt{3+x}+\sqrt{5-x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{2}{(x+1)(\sqrt{3+x}+\sqrt{5-x}) } \Bigg)$

$\displaystyle = \frac{2}{2(\sqrt{4}+\sqrt{4})}$

$\displaystyle = \frac{1}{4}$

$\\$

$\displaystyle \text{Question 21: } \lim \limits_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x }$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(\sqrt{1+x^2}-\sqrt{1-x^2})(\sqrt{1+x^2}+\sqrt{1-x^2})}{x(\sqrt{1+x^2}+\sqrt{1-x^2}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(1+x^2)-(1-x^2)}{x(\sqrt{1+x^2}+\sqrt{1-x^2}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{2x^2}{x(\sqrt{1+x^2}+\sqrt{1-x^2}) } \Bigg)$

$\displaystyle = \frac{2 \times 0}{\sqrt{1+0} + \sqrt{1-0}}$

$\displaystyle = 0$

$\\$

$\displaystyle \text{Question 22: } \lim \limits_{x \to 0} \frac{\sqrt{1+x+x^2}-\sqrt{x+1}}{2x^2 }$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{\sqrt{1+x+x^2}-\sqrt{x+1}}{2x^2 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{1+x+x^2}-\sqrt{x+1}}{2x^2 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(\sqrt{1+x+x^2}-\sqrt{x+1})(\sqrt{1+x+x^2}+\sqrt{x+1})}{2x^2(\sqrt{1+x+x^2}+\sqrt{x+1}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{ (1+x+x^2)-(x+1) }{2x^2(\sqrt{1+x+x^2}+\sqrt{x+1}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{ x^2 }{2x^2(\sqrt{1+x+x^2}+\sqrt{x+1}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{ 1 }{2(\sqrt{1+x+x^2}+\sqrt{x+1}) } \Bigg)$

$\displaystyle = \frac{ 1 }{2(\sqrt{1+0+0^2}+\sqrt{0+1}) }$

$\displaystyle = \frac{1}{4}$

$\\$

$\displaystyle \text{Question 23: } \lim \limits_{x \to 4} \frac{2-\sqrt{x}}{4-x }$

Answer:

$\displaystyle \text{When } x = 4, \text{ the expression } \frac{2-\sqrt{x}}{4-x } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 4} \Bigg( \frac{2-\sqrt{x}}{4-x } \Bigg)$

$\displaystyle = \lim \limits_{x \to 4} \Bigg( \frac{2-\sqrt{x}}{2^2-(\sqrt{x})^2 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 4} \Bigg( \frac{2-\sqrt{x}}{(2-\sqrt{x})(2+\sqrt{x}) } \Bigg)$

$\displaystyle = \frac{1}{2+\sqrt{4}}$

$\displaystyle = \frac{1}{2+2}$

$\displaystyle =\frac{1}{4}$

$\\$

$\displaystyle \text{Question 24: } \lim \limits_{x \to a} \frac{x-a}{ \sqrt{x} - \sqrt{a}}$

Answer:

$\displaystyle \text{When } x = a, \text{ the expression } \frac{x-a}{ \sqrt{x} - \sqrt{a}} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to a} \Bigg( \frac{x-a}{ \sqrt{x} - \sqrt{a}} \Bigg)$

$\displaystyle = \lim \limits_{x \to a} \Bigg( \frac{(\sqrt{x})^2-(\sqrt{a})^2}{ \sqrt{x} - \sqrt{a}} \Bigg)$

$\displaystyle = \lim \limits_{x \to a} \Bigg( \frac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{ \sqrt{x} - \sqrt{a}} \Bigg)$

$\displaystyle = \sqrt{a}+\sqrt{a}$

$\displaystyle = 2\sqrt{a}$

$\\$

$\displaystyle \text{Question 25: } \lim \limits_{x \to 0} \frac{\sqrt{1+3x}-\sqrt{1-3x}}{x }$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{\sqrt{1+3x}-\sqrt{1-3x}}{x } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{1+3x}-\sqrt{1-3x}}{x } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(\sqrt{1+3x}-\sqrt{1-3x})(\sqrt{1+3x}+\sqrt{1-3x})}{x(\sqrt{1+3x}+\sqrt{1-3x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(1+3x)-(1-3x)}{x(\sqrt{1+3x}+\sqrt{1-3x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{6x}{x(\sqrt{1+3x}+\sqrt{1-3x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{6}{(\sqrt{1+3x}+\sqrt{1-3x}) } \Bigg)$

$\displaystyle = \frac{6}{\sqrt{1}+\sqrt{1}}$

$\displaystyle = \frac{6}{2}$

$\displaystyle = 3$

$\\$

$\displaystyle \text{Question 26: } \lim \limits_{x \to 0} \frac{\sqrt{2-x}-\sqrt{2+x}}{x }$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{\sqrt{2-x}-\sqrt{2+x}}{x }\text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{2-x}-\sqrt{2+x}}{x } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(\sqrt{2-x}-\sqrt{2+x})(\sqrt{2-x}+\sqrt{2+x})}{x(\sqrt{2-x}+\sqrt{2+x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(2-x)-(2+x)}{x(\sqrt{2-x}+\sqrt{2+x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{-2x}{x(\sqrt{2-x}+\sqrt{2+x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{-2}{(\sqrt{2-x}+\sqrt{2+x}) } \Bigg)$

$\displaystyle = \frac{-2}{2\sqrt{2}}$

$\displaystyle = \frac{-1}{\sqrt{2}}$

$\\$

$\displaystyle \text{Question 27: } \lim \limits_{x \to 1} \frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1 }$

Answer:

$\displaystyle \text{When } x = 1, \text{ the expression } \frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(\sqrt{3+x}-\sqrt{5-x})(\sqrt{3+x}+\sqrt{5-x})}{(x^2-1)(\sqrt{3+x}+\sqrt{5-x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(3+x)-(5-x)}{(x-1)(x+1)(\sqrt{3+x}+\sqrt{5-x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{2(x-1)}{(x-1)(x+1)(\sqrt{3+x}+\sqrt{5-x}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{2}{(x+1)(\sqrt{3+x}+\sqrt{5-x}) } \Bigg)$

$\displaystyle = \frac{2}{(1+1)(\sqrt{3+1}+\sqrt{5-1}) }$

$\displaystyle = \frac{1}{4}$

$\\$

$\displaystyle \text{Question 28: } \lim \limits_{x \to 1} \frac{(2x-3)(\sqrt{x}-1)}{3x^2+3x-6 }$

Answer:

$\displaystyle \text{When } x = 1, \text{ the expression } \frac{(2x-3)(\sqrt{x}-1)}{3x^2+3x-6 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{(2x-3)(\sqrt{x}-1)}{3x^2+3x-6 } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(2x-3)(\sqrt{x}-1)}{3(x^2+x-2) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(2x-3)(\sqrt{x}-1)}{3(x-1)(x+2) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(2x-3)(\sqrt{x}-1)}{3(\sqrt{x}-1)(\sqrt{x}+1)(x+2) } \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{(2x-3)}{3(\sqrt{x}+1)(x+2) } \Bigg)$

$\displaystyle = \frac{(2-3)}{3(\sqrt{1}+1)(1+2) }$

$\displaystyle = \frac{-1}{18}$

$\\$

$\displaystyle \text{Question 29: } \lim \limits_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x} }$

Answer:

$\displaystyle \text{When } x = 0, \text{ the expression } \frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x} } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 0} \Bigg( \frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x} } \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(\sqrt{1+x^2}-\sqrt{1+x})(\sqrt{1+x^2}+\sqrt{1+x})}{(\sqrt{1+x^2}+\sqrt{1+x}) } \times \frac{(\sqrt{1+x^3}+\sqrt{1+x})}{(\sqrt{1+x^3}-\sqrt{1+x})(\sqrt{1+x^3}+\sqrt{1+x}) }\Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{(1+x^2)-(1-x)}{(1+x^3)-(1+x)} \times \frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{x^2-x}{x^3-x} \times \frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}} \Bigg)$

$\displaystyle = \lim \limits_{x \to 0} \Bigg( \frac{x(x-1)}{x(x-1)(x+1)} \times \frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}} \Bigg)$

$\displaystyle = \frac{\sqrt{1+0}+\sqrt{1+0}}{(0+1)(\sqrt{1+0}+\sqrt{1+0})}$

$\displaystyle = \frac{2}{2}$

$\displaystyle = 1$

$\\$

$\displaystyle \text{Question 30: } \lim \limits_{x \to 1} \frac{x^2-\sqrt{x}}{\sqrt{x}-1}$

Answer:

$\displaystyle \text{When } x = 1, \text{ the expression } \frac{x^2-\sqrt{x}}{\sqrt{x}-1} \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to 1} \Bigg( \frac{x^2-\sqrt{x}}{\sqrt{x}-1} \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{\sqrt{x}( x\sqrt{x} - 1 ) }{(\sqrt{x}-1)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{\sqrt{x}( (\sqrt{x})^3 - 1 ) }{(\sqrt{x}-1)} \Bigg)$

$\displaystyle = \lim \limits_{x \to 1} \Bigg( \frac{\sqrt{x}( \sqrt{x}-1 ) ( x+\sqrt{x}+1 ) }{(\sqrt{x}-1)} \Bigg)$

$\displaystyle = 1 ( 1 + 1 + 1)$

$\displaystyle =3$

$\\$

$\displaystyle \text{Question 31: } \lim \limits_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h } , x \neq 0$

Answer:

$\displaystyle \text{When } h = 0, \text{ the expression } \frac{\sqrt{x+h}-\sqrt{x}}{h } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{h \to 0} \Bigg( \frac{\sqrt{x+h}-\sqrt{x}}{h } \Bigg)$

$\displaystyle = \lim \limits_{h \to 0} \Bigg( \frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{h \to 0} \Bigg( \frac{x+h - x}{h(\sqrt{x+h}+\sqrt{x}) } \Bigg)$

$\displaystyle = \lim \limits_{h \to 0} \Bigg( \frac{h}{h(\sqrt{x+h}+\sqrt{x}) } \Bigg)$

$\displaystyle = \frac{1}{\sqrt{x}+\sqrt{x}}$

$\displaystyle = \frac{1}{2\sqrt{x}}$

$\\$

$\displaystyle \text{Question 32: } \lim \limits_{x \to \sqrt{10}} \frac{\sqrt{(7+2x)}-(\sqrt{5}+\sqrt{2})}{x^2-10 }$

Answer:

$\displaystyle \text{When } x = \sqrt{10}, \text{ the expression } \frac{\sqrt{(7+2x)}-(\sqrt{5}+\sqrt{2})}{x^2-10 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to \sqrt{10}} \Bigg( \frac{\sqrt{(7+2x)}-(\sqrt{5}+\sqrt{2})}{x^2-10 } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{10}} \Bigg( \frac{\sqrt{(7+2x) } - (\sqrt{ (\sqrt{5}+\sqrt{2}) })^2 }{x^2-10 } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{10}} \Bigg( \frac{\sqrt{(7+2x)}- (\sqrt{ 7+2\sqrt{10}}) }{(x-\sqrt{10})(x+\sqrt{10}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{10}} \Bigg( \frac{(\sqrt{(7+2x)}- (\sqrt{ 7+2\sqrt{10}})) (\sqrt{(7+2x)}+ (\sqrt{ 7+2\sqrt{10}})) }{(x-\sqrt{10})(x+\sqrt{10})(\sqrt{(7+2x)}+ (\sqrt{ 7+2\sqrt{10}})) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{10}} \Bigg( \frac{ 2(x-\sqrt{10}) }{(x-\sqrt{10})(x+\sqrt{10})((7+2x)+ (\sqrt{ 7+2\sqrt{10}})) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{10}} \Bigg( \frac{ 2 }{(x+\sqrt{10})(\sqrt{(7+2x)}+ (\sqrt{ 7+2\sqrt{10}})) } \Bigg)$

$\displaystyle = \frac{ 2 }{(\sqrt{10}+\sqrt{10})((\sqrt{7+2\sqrt{10}})+ (\sqrt{ 7+2\sqrt{10}})) }$

$\displaystyle = \frac{2}{(2\sqrt{10}) \times 2 (\sqrt{ 7+2\sqrt{10}}) }$

$\displaystyle = \frac{1}{(2\sqrt{10}) \times \sqrt{ (\sqrt{5}+\sqrt{2} )^2 } }$

$\displaystyle = \frac{1}{(2\sqrt{10}) \times (\sqrt{5}+\sqrt{2} ) }$

$\displaystyle = \frac{(\sqrt{5}-\sqrt{2} )}{(2\sqrt{10}) \times (\sqrt{5}+\sqrt{2} )(\sqrt{5}-\sqrt{2} ) }$

$\displaystyle = \frac{(\sqrt{5}-\sqrt{2} )}{ 6\sqrt{10} }$

$\\$

$\displaystyle \text{Question 33: } \lim \limits_{x \to \sqrt{6}} \frac{\sqrt{(5+2x)}-(\sqrt{3}+\sqrt{2})}{x^2-6 }$

Answer:

$\displaystyle \text{When } x = \sqrt{6}, \text{ the expression } \frac{\sqrt{(5+2x)}-(\sqrt{3}+\sqrt{2})}{x^2-6 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to \sqrt{6}} \Bigg( \frac{\sqrt{(5+2x)}-(\sqrt{3}+\sqrt{2})}{x^2-6 } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{6}} \Bigg( \frac{\sqrt{(5+2x)}-(\sqrt{3}+\sqrt{2})}{(x-\sqrt{6})(x+\sqrt{6}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{6}} \Bigg( \frac{\sqrt{(5+2x)}-(\sqrt{(\sqrt{3}+\sqrt{2})^2}}{(x-\sqrt{6})(x+\sqrt{6}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{6}} \Bigg( \frac{\sqrt{(5+2x)}-\sqrt{5+2\sqrt{6}} }{(x-\sqrt{6})(x+\sqrt{6}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{6}} \Bigg( \frac{(\sqrt{(5+2x)}-\sqrt{5+2\sqrt{6}} )(\sqrt{(5+2x)}+\sqrt{5+2\sqrt{6}} ) }{(x-\sqrt{6})(x+\sqrt{6}) (\sqrt{(5+2x)}+\sqrt{5+2\sqrt{6}} )} \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{6}} \Bigg( \frac{ 2(x-\sqrt{6}) }{(x-\sqrt{6})(x+\sqrt{6}) (\sqrt{(5+2x)}+\sqrt{5+2\sqrt{6}} )} \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{6}} \Bigg( \frac{ 2 }{(x+\sqrt{6}) (\sqrt{(5+2x)}+\sqrt{5+2\sqrt{6}} )} \Bigg)$

$\displaystyle = \frac{1}{2\sqrt{6} (\sqrt{3}+\sqrt{2})}$

$\displaystyle = \frac{ (\sqrt{3}-\sqrt{2})}{2\sqrt{6} (\sqrt{3}+\sqrt{2}) (\sqrt{3}-\sqrt{2})}$

$\displaystyle = \frac{ (\sqrt{3}-\sqrt{2})}{2\sqrt{6} (3-2)}$

$\displaystyle = \frac{ (\sqrt{3}-\sqrt{2})}{2\sqrt{6}}$

$\\$

$\displaystyle \text{Question 34: } \lim \limits_{x \to \sqrt{2}} \frac{\sqrt{(3+2x)}-(\sqrt{2}+1)}{x^2-2 }$

Answer:

$\displaystyle \text{When } x = \sqrt{2}, \text{ the expression } \frac{\sqrt{(3+2x)}-(\sqrt{2}+1)}{x^2-2 } \text{ assumes the form } \frac{0}{0}.$

Rationalizing the numerator, we get

$\displaystyle \lim \limits_{x \to \sqrt{2}} \Bigg( \frac{\sqrt{(3+2x)}-(\sqrt{2}+1)}{x^2-2 } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{2}} \Bigg( \frac{\sqrt{(3+2x)}-\sqrt{(\sqrt{2}+1)^2} }{(x-\sqrt{2})(x+\sqrt{2}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{2}} \Bigg( \frac{\sqrt{(3+2x)}-\sqrt{ 3+2\sqrt{2} } }{(x-\sqrt{2})(x+\sqrt{2}) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{2}} \Bigg( \frac{(\sqrt{(3+2x)}-\sqrt{ 3+2\sqrt{2} } ) (\sqrt{(3+2x)}+\sqrt{ 3+2\sqrt{2} } )}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{(3+2x)}+\sqrt{ 3+2\sqrt{2} } ) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{2}} \Bigg( \frac{ 2(x-\sqrt{2}) }{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{(3+2x)}+\sqrt{ 3+2\sqrt{2} } ) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{2}} \Bigg( \frac{ 2 }{(x+\sqrt{2})(\sqrt{(3+2x)}+\sqrt{ 3+2\sqrt{2} } ) } \Bigg)$

$\displaystyle = \lim \limits_{x \to \sqrt{2}} \Bigg( \frac{ 2 }{(\sqrt{2}+\sqrt{2})(\sqrt{(3+2\sqrt{2})}+\sqrt{ 3+2\sqrt{2} } ) } \Bigg)$

$\displaystyle = \frac{2}{2\sqrt{2} (2\sqrt{3+2\sqrt{2}}) }$

$\displaystyle = \frac{1}{2\sqrt{2} (\sqrt{ (\sqrt{2}+1)^2 } ) }$

$\displaystyle = \frac{1}{2\sqrt{2} (\sqrt{2}+1) }$

$\displaystyle = \frac{(\sqrt{2}-1)}{2\sqrt{2} (\sqrt{2}+1)(\sqrt{2}-1) }$

$\displaystyle = \frac{(\sqrt{2}-1)}{2\sqrt{2} }$

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Class 11: Limits – Exercise 29.4

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