Class 11: Limits – Exercise 29.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Evaluate the following limits:

$\displaystyle \text{Question 1: } \lim \limits_{x \to -5} \frac{2x^2+9x-5}{x+5 }$

Answer:

$\displaystyle \text{When } x = -5, \text{ the expression } \frac{2x^2+9x-5}{x+5 } \text{ assumes the form } \frac{0}{0}.$

So, $(x+5)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 5} \frac{2x^2+9x-5}{x+5 }$

$\displaystyle = \lim \limits_{x \to 5} \frac{2x^2+10x - x-5}{x+5 }$

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

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

$\displaystyle = \lim \limits_{x \to 5} (2x-1)$

$\displaystyle = 2(-5) - 1$

$\displaystyle = - 11$

$\\$

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

Answer:

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

So, $(x-3)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 3} \frac{ x^2-4x+3 }{x^2-2x-3 }$

$\displaystyle = \lim \limits_{x \to 3} \frac{ x^2-x-3x+3 }{x^2-3x+x-3 }$

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

$\displaystyle = \lim \limits_{x \to 3} \frac{ x-1 }{x+1 }$

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

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

$\\$

$\displaystyle \text{Question 3: } \lim \limits_{x \to 3} \frac{x^4-81 }{ x^2-9}$

Answer:

$\displaystyle \text{When } x = 3, \text{ the expression } \frac{x^4-81 }{ x^2-9} \text{ assumes the form } \frac{0}{0}.$

So, $(x-3)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 3} \frac{x^4-81 }{ x^2-9}$

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

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

$\displaystyle =\lim \limits_{x \to 3} (x^2-9)$

$\displaystyle = 3^2+9$

$\displaystyle = 18$

$\\$

$\displaystyle \text{Question 4: } \lim \limits_{x \to 2} \frac{ x^3-8 }{x^2-4 }$

Answer:

$\displaystyle \text{When } x = 2, \text{ the expression } \frac{ x^3-8 }{x^2-4 } \text{ assumes the form } \frac{0}{0}.$

So, $(x-2)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 2} \frac{ x^3-8 }{x^2-4 }$

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

$\displaystyle = \lim \limits_{x \to 2} \frac{ x^2+2x+4 }{x+2}$

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

$\displaystyle = \frac{12}{4}$

$\displaystyle = 3$

$\\$

$\displaystyle \text{Question 5: } \lim \limits_{x \to-\frac{1}{2} } \frac{8x^3+1 }{2x+1 }$

Answer:

$\displaystyle \text{When } x = -\frac{1}{2}, \text{ the expression } \frac{8x^3+1 }{2x+1 } \text{ assumes the form } \frac{0}{0}.$

So, $\displaystyle (x+\frac{1}{2})$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to-\frac{1}{2} } \frac{8x^3+1 }{2x+1 }$

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

$\displaystyle = \lim \limits_{x \to-\frac{1}{2} } \frac{(2x+1)[ (2x)^2-2x \times 1 + 1^2] }{2x+1 }$

$\displaystyle = \lim \limits_{x \to-\frac{1}{2} } [ (2x)^2-2x \times 1 + 1^2]$

$\displaystyle = \Bigg( 2 \times \frac{-1}{2} \Bigg) ^2 - 2 \times \frac{-1}{2} + 1$

$\displaystyle = 1 + 1 + 1$

$\displaystyle = 3$

$\\$

$\displaystyle \text{Question 6: } \lim \limits_{x \to 4} \frac{x^2-7x+12 }{x^2-3x-4 }$

Answer:

$\displaystyle \text{When } x = 4, \text{ the expression } \frac{x^2-7x+12 }{x^2-3x-4 } \text{ assumes the form } \frac{0}{0}.$

So, $(x-4)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 4} \frac{x^2-7x+12 }{x^2-3x-4 }$

$\displaystyle = \lim \limits_{x \to 4} \frac{x^2-3x-4x+12 }{x^2-4x+x-4 }$

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

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

$\displaystyle = \frac{4-3}{4+1}$

$\displaystyle = \frac{1}{5}$

$\\$

$\displaystyle \text{Question 7: } \lim \limits_{x \to 2} \frac{x^4-16 }{x-2 }$

Answer:

$\displaystyle \text{When } x = 2, \text{ the expression } \frac{x^4-16 }{x-2 } \text{ assumes the form } \frac{0}{0}.$

So, $(x-2)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 2} \frac{x^4-16 }{x-2 }$

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

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

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

$\displaystyle = (2+2_)(2^2+4)$

$\displaystyle = 32$

$\\$

$\displaystyle \text{Question 8: } \lim \limits_{x \to 5} \frac{ x^2-9x+20 }{x^2-6x+5 }$

Answer:

$\displaystyle \text{When } x = 5, \text{ the expression } \frac{ x^2-9x+20 }{x^2-6x+5 } \text{ assumes the form } \frac{0}{0}.$

So, $(x-5)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 5} \frac{ x^2-9x+20 }{x^2-6x+5 }$

$\displaystyle = \lim \limits_{x \to 5} \frac{ x^2-4x-5x+20 }{x^2-x-5x+5 }$

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

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

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

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

$\\$

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

Answer:

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

So, $(x+1)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to -1} \frac{ x^3+1 }{ x+1}$

$\displaystyle = \lim \limits_{x \to -1} \frac{ x^3+1^3 }{ x+1}$

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

$\displaystyle = (-1)^2-(-1)+1$

$\displaystyle = 1+1+1$

$\displaystyle = 3$

$\\$

$\displaystyle \text{Question 10: } \lim \limits_{x \to 5 } \frac{x^3-125 }{x^2-7x+10 }$

Answer:

$\displaystyle \text{When } x = 5, \text{ the expression } \frac{x^3-125 }{x^2-7x+10 } \text{ assumes the form } \frac{0}{0}.$

So, $(x-5)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 5 } \frac{x^3-125 }{x^2-7x+10 }$

$\displaystyle = \lim \limits_{x \to 5 } \frac{x^3-5^3 }{x^2-2x-5x+10 }$

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

$\displaystyle = \lim \limits_{x \to 5 } \frac{(x-5)(x^2+5x+25) }{(x-2)(x-5) }$

$\displaystyle = \frac{5^2+ 5 \times 5 + 25}{5-2}$

$\displaystyle = \frac{75}{3}$

$\displaystyle = 25$

$\\$

$\displaystyle \text{Question 11: } \lim \limits_{x \to \sqrt{2}} \frac{ x^2-2 }{ x^2+ \sqrt{2} x - 4}$

Answer:

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

So, $(x-\sqrt{2})$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to \sqrt{2}} \frac{ x^2-2 }{ x^2+ \sqrt{2} x - 4}$

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

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

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

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

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

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

$\\$

$\displaystyle \text{Question 12: } \lim \limits_{x \to \sqrt{3} } \frac{ x^2-3 }{x^2 + 3\sqrt{3}x- 12 }$

Answer:

$\displaystyle \text{When } x = \sqrt{3}, \text{ the expression } \frac{ x^2-3 }{x^2 + 3\sqrt{3}x- 12 } \text{ assumes the form } \frac{0}{0}.$

So, $(x-\sqrt{3})$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to \sqrt{3} } \frac{ x^2-3 }{x^2 + 3\sqrt{3}x- 12 }$

$\displaystyle = \lim \limits_{x \to \sqrt{3} } \frac{ x^2-(\sqrt{3})^2 }{x^2 + 4\sqrt{3}x - \sqrt{3}x- 12 }$

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

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

$\displaystyle = \frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}+ 4 \sqrt{3}}$

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

$\\$

$\displaystyle \text{Question 13: } \lim \limits_{x \to \sqrt{3} } \frac{ x^4 - 9 }{ x^2 + 4 \sqrt{3} x - 15}$

Answer:

$\displaystyle \text{When } x = \sqrt{3}, \text{ the expression } \frac{ x^4 - 9 }{ x^2 + 4 \sqrt{3} x - 15} \text{ assumes the form } \frac{0}{0}.$

So, $(x-\sqrt{3})$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to \sqrt{3} } \frac{ x^4 - 9 }{ x^2 + 4 \sqrt{3} x - 15}$

$\displaystyle = \lim \limits_{x \to \sqrt{3} } \frac{ (x^2)^2 - 3^2 }{ x^2 + 5 \sqrt{3} x - \sqrt{3} x - 15}$

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

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

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

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

$\displaystyle = \frac{2\sqrt{3} \times 6}{6\sqrt{3}}$

$\displaystyle = 2$

$\\$

$\displaystyle \text{Question 14: } \lim \limits_{x \to 2} \Bigg( \frac{ x }{x-2 } - \frac{4}{x^2-2x} \Bigg)$

Answer:

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

So, $(x-2)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

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

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

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

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

$\displaystyle = 2$

$\\$

$\displaystyle \text{Question 15: } \lim \limits_{x \to 1 } \Bigg( \frac{ 1 }{x^2+x-2 } - \frac{x}{x^3-1} \Bigg)$

Answer:

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

So, $(x-1)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

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

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

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

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

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

$\displaystyle = \lim \limits_{x \to 1 } \frac{ -(x-1) }{[ x(x+2) - 1 ( x+2 )] ( x^2 + x + 1) }$

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

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

$\displaystyle = \frac{-1}{9}$

$\\$

$\displaystyle \text{Question 16: } \lim \limits_{x \to 3 } \Bigg( \frac{ 1 }{x-3 } - \frac{2}{x^2-4x+3} \Bigg)$

Answer:

$\displaystyle \text{When } x = 3, \text{ the expression } \Bigg( \frac{ 1 }{x-3 } - \frac{2}{x^2-4x+3} \Bigg) \text{ assumes the form } \frac{0}{0}.$

So, $(x-3)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

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

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

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

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

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

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

$\\$

$\displaystyle \text{Question 17: } \lim \limits_{x \to 2 } \Bigg( \frac{ 1 }{x-2 } - \frac{2}{x^2-2x} \Bigg)$

Answer:

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

So, $(x+2)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

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

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

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

$\\$

$\displaystyle \text{Question 18: } \lim \limits_{x \to \frac{1}{4} } \frac{ 4x-1 }{2\sqrt{x}-1 }$

Answer:

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

So, $(x-\frac{1}{4})$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to \frac{1}{4} } \frac{ 4x-1 }{2\sqrt{x}-1 }$

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

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

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

$\displaystyle = 2$

$\\$

$\displaystyle \text{Question 19: } \lim \limits_{x \to 4 } \frac{ x^2-16 }{\sqrt{x} -2 }$

Answer:

$\displaystyle \text{When } x = 4, \text{ the expression } \frac{ x^2-16 }{\sqrt{x} -2 } \text{ assumes the form } \frac{0}{0}.$

So, $(x-4)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 4 } \frac{ x^2-16 }{\sqrt{x} -2 }$

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

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

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

$\displaystyle = (2+2)(4+4)$

$\displaystyle = 32$

$\\$

$\displaystyle \text{Question 20: } \lim \limits_{x \to 0} \frac{ (a+x)^2-a^2 }{x }$

Answer:

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

So, $(x-0)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

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

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

$\displaystyle = 0+2a$

$\displaystyle = 2a$

$\\$

$\displaystyle \text{Question 21: } \lim \limits_{x \to 2} \Bigg( \frac{ 1 }{x-2 } - \frac{4}{x^3-2x^2} \Bigg)$

Answer:

$\displaystyle \text{When } x = 2, \text{ the expression } \Bigg( \frac{ 1 }{x-2 } - \frac{4}{x^3-2x^2} \Bigg) \text{ assumes the form } \frac{0}{0}.$

So, $(x-2)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

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

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

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

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

$\displaystyle = 1$

$\\$

$\displaystyle \text{Question 22: } \lim \limits_{x \to 3} \Bigg( \frac{ 1 }{x-3 } - \frac{3}{x^2-3x} \Bigg)$

Answer:

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

So, $(x-3)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

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

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

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

$\\$

$\displaystyle \text{Question 23: } \lim \limits_{x \to 1} \Bigg( \frac{ 1 }{x-1 } - \frac{2}{x^2-1} \Bigg)$

Answer:

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

So, $(x-1)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

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

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

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

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

$\\$

$\displaystyle \text{Question 24: } \lim \limits_{x \to 3} (x^2-9)\Bigg( \frac{ 1 }{x+3 } + \frac{1}{x-3} \Bigg)$

Answer:

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

So, $(x-3)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 3} (x^2-9)\Bigg( \frac{ 1 }{x+3 } + \frac{1}{x-3} \Bigg)$

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

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

$\displaystyle = \lim \limits_{x \to 3} 2x$

$\displaystyle = 2 \times 3$

$\displaystyle = 6$

$\\$

$\displaystyle \text{Question 25: } \lim \limits_{x \to 1} \frac{ x^4-3x^3+2 }{x^3-5x^2+3x+1 }$

Answer:

$\displaystyle \text{When } x = 1, \text{ the expression } \frac{ x^4-3x^3+2 }{x^3-5x^2+3x+1 } \text{ assumes the form } \frac{0}{0}.$

So, $(x-1)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 1} \frac{ x^4-3x^3+2 }{x^3-5x^2+3x+1 }$

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

$\displaystyle = \frac{1^3- 2(1)^2-2(1) - 2}{(1)^2-4\times1-1}$

$\displaystyle = \frac{-5}{-4}$

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

$\\$

$\displaystyle \text{Question 26: } \lim \limits_{x \to 2} \frac{ x^3+3x^2-9x-2 }{ x^3-x-6}$

Answer:

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

So, $(x-2)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

$\displaystyle = \lim \limits_{x \to 2} \frac{ x^2+5x+1 }{x^2+2x+3}$

$\displaystyle = \frac{(2)^2+5 \times 2 + 1}{(2)^2+2 \times 2 + 3}$

$\displaystyle = \frac{4 + 10 + 1}{4 + 4 + 3}$

$\displaystyle = \frac{15}{11}$

$\\$

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

Answer:

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

So, $(x-1)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 1} \frac{1-x^{-\frac{1}{3}} }{1-x^{-\frac{2}{3}}}$

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

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

$\displaystyle = \lim \limits_{x \to 1} \frac{1 }{1+x^{-\frac{1}{3}}}$

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

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

$\\$

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

Answer:

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

So, $(x-3)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to 3} \frac{x^2-x-6 }{x^3-3x^2+x-3 }$

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

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

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

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

$\displaystyle = \frac{5}{10}$

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

$\\$

$\displaystyle \text{Question 29: } \lim \limits_{x \to -2} \frac{x^3+x^2+4x+12 }{x^3-3x+2 }$

Answer:

$\displaystyle \text{When } x = -2, \text{ the expression } \frac{x^3+x^2+4x+12 }{x^3-3x+2 } \text{ assumes the form } \frac{0}{0}.$

So, $(x+2)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

$\displaystyle \lim \limits_{x \to -2} \frac{x^3+x^2+4x+12 }{x^3-3x+2 }$

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

$\displaystyle = \frac{2^2-(-2)+6}{(-2)^2-2(-2)+1}$

$\displaystyle = \frac{4+2+6}{4+4+1}$

$\displaystyle = \frac{12}{9}$

$\displaystyle = \frac{4}{3}$

$\\$

$\displaystyle \text{Question 30: } \lim \limits_{x \to 1} \frac{x^3+3x^2-6x+2 }{x^3+3x^2-3x-1 }$

Answer:

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

So, $(x-1)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

$\displaystyle = \frac{1^2+4\times 1 - 2}{1^2+4 \times 1 + 1}$

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

$\displaystyle = \frac{3}{6}$

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

$\\$

$\displaystyle \text{Question 31: } \lim \limits_{x \to 2} \Bigg( \frac{ 1 }{x-2 } - \frac{2(2x-3)}{x^3-3x^2+2x} \Bigg)$

Answer:

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

So, $(x-2)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

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

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

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

$\displaystyle = \lim \limits_{x \to 2} \Bigg( \frac{x^2-x-4x+6 }{x(x-1)(x-2) } \Bigg)$

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

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

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

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

$\\$

$\displaystyle \text{Question 32: } \lim \limits_{x \to 1} \frac{ \sqrt{x^2-1} + \sqrt{x-1} }{\sqrt{x^2-1} }, x> 1$

Answer:

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

So, $(x-1)$ is a common factor in numerator and denominator. Factorizing the numerator and denominator, we get

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

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

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

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

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

$\\$

$\displaystyle \text{Question 33: } \lim \limits_{x \to 1} \Bigg( \frac{ x-2 }{x^2-x } + \frac{1}{x^3-3x^2+2x} \Bigg)$

Answer:

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