Note:

\displaystyle \lim \limits_{x \to 0 } \Bigg( \frac{a^x-1}{x} \Bigg) = \log a  

Evaluate the following limits: 

\displaystyle \text{Question 1: } \lim \limits_{x \to 0 } \ \frac{5^x-1}{\sqrt{4+x} - 2} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{5^x-1}{\sqrt{4+x} - 2} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{(5^x-1)( \sqrt{4+x}+2)}{(\sqrt{4+x} - 2)( \sqrt{4+x}+2)} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{(5^x-1)( \sqrt{4+x}+2)}{4+x-4} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg( \frac{5^x-1}{x}\Bigg) ( \sqrt{4+x}+2)

\displaystyle = \log 5 \times ( \sqrt{4+0}+2) = 4 \log 5

\\

\displaystyle \text{Question 2: } \lim \limits_{x \to 0 } \ \frac{ \log(1+x)}{3^x-1} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{ \log(1+x)}{3^x-1} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{ \log(1+x)}{x \cdot \Big( \frac{3^x-1}{x} \Big) } 

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

\\

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

Answer:

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

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

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

\displaystyle = \lim \limits_{x \to 0 } \ \frac{ \Big( a^{\frac{x}{2}} - \frac{1}{a^{\frac{x}{2}}} \Big)^2 }{x^2}

\displaystyle = \lim \limits_{x \to 0 } \ \frac{(a^x-1)^2}{x^2} \times \frac{1}{ \Big( a^{\frac{x}{2}} \Big)^2}

\displaystyle = \lim \limits_{x \to 0 } \Bigg( \frac{a^x - 1}{x} \Bigg)^2 \times \frac{1}{a^x}

\displaystyle = \frac{(\log a)^2}{a^0} = (\log a)^2

\\

\displaystyle \text{Question 4: } \lim \limits_{x \to 0 } \ \frac{a^{mx}-1}{b^{nx}-1}, n \neq 0 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{a^{mx}-1}{b^{nx}-1}

\displaystyle = \lim \limits_{x \to 0 } \Bigg( \frac{a^{mx} -1}{mx} \times \frac{mx}{\frac{b^{nx}-1}{nx} \times nx } \Bigg)

\displaystyle = \frac{\log a}{\log b} \times \frac{m}{n} = \frac{m}{n} \frac{\log a}{\log b}

\\

\displaystyle \text{Question 5: } \lim \limits_{x \to 0 } \ \frac{a^x+b^x-2}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{a^x+b^x-2}{x} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{a^x-1+b^x-1}{x} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg( \frac{a^x-1}{x} +\frac{b^x-1}{x} \Bigg) 

\displaystyle =\log a + \log b

\displaystyle = \log (ab)

\\  

\displaystyle \text{Question 6: } \lim \limits_{x \to 0 } \ \frac{9^x- 2 \cdot 6^x + 4^x}{x^2} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{9^x- 2 \cdot 6^x + 4^x}{x^2} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{(3^x)^2- 2 \cdot 3^x \cdot 2^x + (2^x)^2}{x^2} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{(3^x-2^x)^2}{x^2}

\displaystyle = \lim \limits_{x \to 0 } \Bigg[  \Bigg( \frac{3^x-2^x}{2^x}  \Bigg)^2 \times \frac{(2^x)^2}{x^2}  \Bigg]

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

\displaystyle = \Bigg[ \log \Big( \frac{3}{2} \Big) \Bigg]^2 \times 2^0

\displaystyle = \Bigg[ \log \Big( \frac{3}{2} \Big) \Bigg]^2

\\

\displaystyle \text{Question 7: } \lim \limits_{x \to 0 } \ \frac{8^x-4^x - 2^x + 1}{x^2} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{8^x-4^x - 2^x + 1}{x^2} 

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

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

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

\displaystyle = \lim \limits_{x \to 0 } \ \frac{2(2^{2x}-1)}{2x} \times \frac{2^x-1}{x} 

\displaystyle = 2 \log 2 \times \log 2

\displaystyle = \log (2)^2 \times \log 2

\displaystyle = \log 4 \times \log 2

\\

\displaystyle \text{Question 8: } \lim \limits_{x \to 0 } \ \frac{a^{mx} - b^{nx}}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{a^{mx} - b^{nx}}{x} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{a^{mx} - 1 - b^{nx}+1}{x} 

\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \Bigg( \frac{a^{mx} - 1}{mx} \Bigg) \times m - \Bigg( \frac{b^{nx} - 1}{nx} \Bigg) \times n \Bigg]

\displaystyle = m \log a - n \log b

\displaystyle = \log (a)^m - \log (b)^n

\displaystyle = \log \Big( \frac{a^m}{b^n} \Big)

\\

\displaystyle \text{Question 9: } \lim \limits_{x \to 0 } \ \frac{a^x+b^x+c^x-3}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{a^x+b^x+c^x-3}{x} 

\displaystyle = \lim \limits_{x \to 0 } \ \Bigg[ \frac{a^x-1}{x} + \frac{b^x-1}{x} + \frac{c^x-1}{x} \Bigg] 

\displaystyle = \log a + \log b + \log c

\displaystyle = \log (abc)

\\

\displaystyle \text{Question 10: } \lim \limits_{x \to 2 } \ \frac{x-2}{\log_a (x-1)} 

Answer:

\displaystyle \lim \limits_{x \to 2 } \ \frac{x-2}{\log_a (x-1)} 

\displaystyle \text{Let } x=2+h. \text{ When } x \rightarrow 2, h \rightarrow 0

\displaystyle = \lim \limits_{h \to 0 } \ \frac{2+h-2}{\frac{\log (2+h-1)}{\log a} } 

\displaystyle = \log a \lim \limits_{h \to 0 } \Bigg[  \frac{h}{\log (1+h)} \Bigg]

\displaystyle = \log a \times 1 = \log a

\\

\displaystyle \text{Question 11: } \lim \limits_{x \to 0 } \ \frac{5^x+3^x+2^x-3}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \Bigg[ \frac{5^x+3^x+2^x-3}{x}  \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{5^x-1}{x}  + \frac{5^x-1}{x} + \frac{2^x-1}{x}\Bigg]

\displaystyle = \log 5+ \log 3+ \log 2

\displaystyle = \log 30

\\

\displaystyle \text{Question 12: } \lim \limits_{x \to \infty } \ (a^{\frac{1}{x}} -1)x 

Answer:

\displaystyle \lim \limits_{x \to \infty } \ (a^{\frac{1}{x}} -1)x 

\displaystyle \text{Let } y = \frac{1}{x}. \text{ When } x \rightarrow \infty, \text{ then } y \rightarrow 0

\displaystyle = \lim \limits_{y \to 0} \frac{a^y - 1}{y}

\displaystyle = \log a

\\  

\displaystyle \text{Question 13: } \lim \limits_{x \to 0 } \ \frac{a^{mx} - b^{nx}}{\sin kx} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \Bigg[ \frac{a^{mx} - b^{nx}}{\sin kx} \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{(a^{mx}-1) - (b^{nx}-1)}{\sin kx} \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{ m \Big( \frac{a^{mx}-1}{mx} \Big) - n \Big( \frac{b^{nx}-1}{nx} \Big) }{ k \times \frac{\sin kx}{kx} } \Bigg]

\displaystyle = \frac{(m \log a - n \log b)}{k \times 1}

\displaystyle = \frac{1}{k} [ \log (a)^m - \log (b)^n ]

\displaystyle =\frac{1}{k} \log \Big( \frac{a^m}{b^n} \Big)

\\

\displaystyle \text{Question 14: } \lim \limits_{x \to 0 } \ \frac{a^x+b^x-c^x-d^x}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{a^x+b^x-c^x-d^x}{x} 

\displaystyle =\lim \limits_{x \to 0 } \ \frac{(a^x-1)+(b^x-1)- (c^x-1) -(d^x- 1)}{x} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[  \frac{a^x-1}{x} +\frac{b^x-1}{x} - \frac{c^x-1}{x} - \frac{d^x-1}{x} \Bigg]

\displaystyle = \log a + \log b - \log c - \log d

\displaystyle = \log \Big( \frac{ab}{cd} \Big)

\\

\displaystyle \text{Question 15: } \lim \limits_{x \to 0 } \ \frac{e^x-1+\sin x}{x} 

Answer: 

\displaystyle \lim \limits_{x \to 0 } \ \frac{e^x-1+\sin x}{x} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{e^x-1}{x}+\frac{\sin x}{x}  \Bigg]

\displaystyle = 1 + 1 = 2

\\

\displaystyle \text{Question 16: } \lim \limits_{x \to 0 } \ \frac{\sin 2x}{e^x-1} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{\sin 2x}{e^x-1} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{\sin 2x}{ x \Big( \frac{e^x-1}{x} \Big) }  \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{\sin 2x}{ 2x} \times \frac{2}{ \Big( \frac{e^x-1}{x} \Big) }  \Bigg]

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

\\

\displaystyle \text{Question 17: } \lim \limits_{x \to 0 } \ \frac{e^{\sin x} - 1}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{e^{\sin x} - 1}{x} 

\displaystyle =\lim \limits_{x \to 0 } \Bigg[ \frac{e^{\sin x} - 1}{\sin x} \times \frac{\sin x}{x} \Bigg]

\displaystyle \text{When } x \rightarrow 0, \text{ then } \sin x \rightarrow 0

\displaystyle \text{Let } y = \sin x, \text{ then when } x \rightarrow 0, \text{ then } y \rightarrow 0

\displaystyle =\lim \limits_{y \to 0 } \frac{e^{y} - 1}{y} \times \lim \limits_{x \to 0 } \frac{\sin x}{x}

\displaystyle = 1 \times 1 = 1

\\

\displaystyle \text{Question 18: } \lim \limits_{x \to 0 } \ \frac{e^{2x}-e^x}{\sin 2x} 

Answer:

\displaystyle  \lim \limits_{x \to 0 } \ \frac{e^{2x}-e^x}{\sin 2x} 

\displaystyle  = \lim \limits_{x \to 0 } \ \frac{e^x(e^x-1)}{\sin 2x} 

\displaystyle  = \lim \limits_{x \to 0 } \Bigg[ \frac{e^x(e^x-1)}{x} \times \frac{2x}{\sin 2x} \times \frac{1}{2} \Bigg] 

\displaystyle  = e^0 \times 1 \times \frac{1}{1} \times \frac{1}{2} = \frac{1}{2}

\\

\displaystyle \text{Question 19: } \lim \limits_{x \to a } \ \frac{\log x - \log a}{x-a} 

Answer:

\displaystyle \lim \limits_{x \to a } \ \frac{\log x - \log a}{x-a} 

\displaystyle = \lim \limits_{x \to a } \ \frac{\log \Big( \frac{x}{a} \Big) }{a(\frac{x}{a}-1)} 

\displaystyle \text{When } x \rightarrow a, \text{  then } \frac{x}{a} \rightarrow 1

\displaystyle \text{Let } y = \frac{x}{a} -1. \text{ when } x \rightarrow a, \text{ then } y \rightarrow 0

\displaystyle = \lim \limits_{x \to a } \Bigg[  \frac{\log (1+y) }{a \times y} \Bigg] 

\displaystyle =\frac{1}{a} \times 1 = \frac{1}{a}

\\

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

Answer:

\displaystyle  \lim \limits_{x \to 0 } \ \frac{\log (a+x) - \log ( a-x)}{x} 

\displaystyle  = \lim \limits_{x \to 0 } \ \frac{\log \frac{(a+x)}{( a-x)}}{x} 

\displaystyle  = \lim \limits_{x \to 0 } \ \frac{ \Big( \log 1+\frac{(a+x)}{( a-x)} - 1\Big)}{x}   

\displaystyle  = \lim \limits_{x \to 0 } \ \frac{ \Big( \log 1+\frac{(a+x-a+x)}{( a-x)} \Big) }{x} 

\displaystyle  = \lim \limits_{x \to 0 } \ \frac{ \Big( \log 1+\frac{2x}{ a-x} \Big) }{\frac{2x}{a-x} \times \frac{a-x}{2}} 

\displaystyle  \text{When } x \rightarrow 0, \text{ then } \frac{2x}{a-x} \rightarrow 0

\displaystyle  \text{Let } y = \frac{2x}{a-x} 

\displaystyle  = \lim \limits_{y \to 0 } \Bigg[ \frac{\log ( 1+y) }{y} \Bigg] \times  \lim \limits_{x \to 0 } \ \Bigg[ \frac{1}{\frac{a-x}{2}} \Bigg]

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

\\

\displaystyle \text{Question 21: } \lim \limits_{x \to 0 } \ \frac{\log (2+x) + \log 0.5}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{\log (2+x) + \log 0.5}{x} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{\log (1+\frac{x}{2}) }{x} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{\log (1+\frac{x}{2}) }{2 \times \frac{x}{2} } 

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

\\

\displaystyle \text{Question 22: } \lim \limits_{x \to 0 } \ \frac{\log (a+x) - \log a}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{\log (a+x) - \log a}{x} 

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

\displaystyle = \lim \limits_{x \to 0 } \ \frac{\log (1+\frac{x}{a}) }{\frac{x}{a} \times a} 

\displaystyle = \frac{1}{a} \times 1 = \frac{1}{a}

\\

\displaystyle \text{Question 23: } \lim \limits_{x \to 0 } \ \frac{\log ( 3+x) - \log(3-x)}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{\log ( 3+x) - \log(3-x)}{x} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{\log ( \frac{3+x}{3-x}) }{x} 

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

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

\displaystyle = \lim \limits_{x \to 0 } \ \frac{\log ( 1+\frac{2x}{3-x}) }{x} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{\log ( 1+\frac{2x}{3-x}) }{\frac{2x}{3-x} \times \frac{3-x}{2} } 

\displaystyle = \frac{1 \times 2}{3 - 0} = \frac{2}{3}

\\

\displaystyle \text{Question 24: } \lim \limits_{x \to 0 } \ \frac{8^x-2^x}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{8^x-2^x}{x} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{(8^x-1)-(2^x-1)}{x} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{(8^x-1)}{x} - \frac{(2^x-1)}{x} \Bigg]

\displaystyle = \log 8 - \log 2 = \log \Big( \frac{8}{2} \Big) = \log 4

\\

\displaystyle \text{Question 25: } \lim \limits_{x \to 0 } \ \frac{x(2^x-1)}{1 - \cos x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{x(2^x-1)}{1 - \cos x} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[  \Bigg( \frac{2^x-1}{x} \Bigg) \times \frac{x^2}{1- \cos x}   \Bigg] 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[  \Bigg( \frac{2^x-1}{x} \Bigg) \times \frac{x^2}{2 \sin^2 \frac{x}{2}}   \Bigg] 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[  \Bigg( \frac{2^x-1}{x} \Bigg) \times \frac{\frac{x^2}{4} \times 4}{2 \sin^2 \frac{x}{2}}   \Bigg] 

\displaystyle =\log 2 \times \frac{4}{2} \times \frac{1}{1^2} = 2 \log 2 = \log 2^2 = \log 4

\\

\displaystyle \text{Question 26: } \lim \limits_{x \to 0 } \ \frac{\sqrt{1+x}-1}{\log (1+x)} 

Answer:

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

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

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

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

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

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

\\

\displaystyle \text{Question 27: } \lim \limits_{x \to 0 } \ \frac{\log |1+x^3|}{\sin^3 x} 

Answer:

\displaystyle  \lim \limits_{x \to 0 } \ \frac{\log |1+x^3|}{\sin^3 x} 

\displaystyle  = \lim \limits_{x \to 0 } \Bigg[ \frac{\log (1+x^3)}{x^3} \times \frac{x^3}{\sin^3 x}  \Bigg]

\displaystyle  = 1 \times \frac{1}{1^3} = 1

\\

\displaystyle \text{Question 28: } \lim \limits_{x \to \frac{\pi}{2} } \ \frac{a^{\cot x} - a^{\cos x}}{\cot x - \cos x } 

Answer:

\displaystyle \lim \limits_{x \to \frac{\pi}{2} } \ \frac{a^{\cot x} - a^{\cos x}}{\cot x - \cos x } 

\displaystyle = \lim \limits_{x \to \frac{\pi}{2} } a^{\cos x} \Bigg[ \frac{ \frac{a^{\cot x}}{a^{\cos x}} - 1}{\cot x - \cos x }  \Bigg]

\displaystyle = \lim \limits_{x \to \frac{\pi}{2} } a^{\cos x} \Bigg[ \frac{ a^{\cot x - \cos x} - 1}{\cot x - \cos x }  \Bigg]

\displaystyle = 1 \times  \log a = \log a

\\

\displaystyle \text{Question 29: } \lim \limits_{x \to 0 } \ \frac{e^x - 1}{\sqrt{1 - \cos x}} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{e^x - 1}{\sqrt{1 - \cos x}} 

\displaystyle = \lim \limits_{x \to 0 } \ \frac{e^x - 1}{\sqrt{1 - \cos x}}  \times \frac{\sqrt{1 + \cos x}}{\sqrt{1 + \cos x}} \Bigg]

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

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{(e^x - 1)(\sqrt{1 + \cos x})}{| \sin x | }   \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{(e^x - 1)}{x} \times \frac{ (\sqrt{1 + \cos x}) }{ \Big( \frac{| \sin x |}{x} \Big) }   \Bigg]

Left hand limit:

\displaystyle = \lim \limits_{x \to 0^- } \Bigg[ \frac{(e^x - 1)}{x} \times \frac{ (\sqrt{1 + \cos x}) }{ \Big( \frac{| \sin x |}{x} \Big) }   \Bigg]

\displaystyle = \lim \limits_{x \to 0^- } \Bigg[ \frac{(e^x - 1)}{x} \times \frac{ (\sqrt{1 + \cos x}) }{ \Big( \frac{- \sin x}{x} \Big) }   \Bigg]

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

Right hand limit:

\displaystyle = \lim \limits_{x \to 0^+ } \Bigg[ \frac{(e^x - 1)}{x} \times \frac{ (\sqrt{1 + \cos x}) }{ \Big( \frac{| \sin x |}{x} \Big) }   \Bigg]

\displaystyle = \lim \limits_{x \to 0^+ } \Bigg[ \frac{(e^x - 1)}{x} \times \frac{ (\sqrt{1 + \cos x}) }{ \Big( \frac{ \sin x}{x} \Big) }   \Bigg]

\displaystyle = \frac{1 \times \sqrt{2}}{1} =\sqrt{2}

\displaystyle \text{ Left hand limit } \neq \text{ Right hand limit }

Therefore the limit does not exist.

\\

\displaystyle \text{Question 30: } \lim \limits_{x \to 5 } \ \frac{e^x - e^5}{x-5} 

Answer:

\displaystyle \lim \limits_{x \to 5 } \ \frac{e^x - e^5}{x-5} 

\displaystyle = \lim \limits_{x \to 5 } e^5 \Bigg[ \frac{e^{x-5} - 1}{x-5}  \Bigg]

\displaystyle = e^5

\\

\displaystyle \text{Question 31: } \lim \limits_{x \to 0 } \ \frac{e^{x+2}-e^2}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{e^{x+2}-e^2}{x} 

\displaystyle = \lim \limits_{x \to 0 } e^2 \Bigg[ \frac{e^{x}-1}{x}  \Bigg]

\displaystyle = e^2

\\

\displaystyle \text{Question 32: } \lim \limits_{x \to \frac{\pi}{2} } \ \frac{e^{\cos x}-1}{\cos x} 

Answer:

\displaystyle \lim \limits_{x \to \frac{\pi}{2} } \ \frac{e^{\cos x}-1}{\cos x} 

\displaystyle \text{If } x = \frac{\pi}{2}, \text{ then } \cos x \rightarrow 0

\displaystyle \text{Let } y = \cos x

\displaystyle = \lim \limits_{y \to 0 } \ \frac{e^{y}-1}{y} 

\displaystyle = 1

\\

\displaystyle \text{Question 33: } \lim \limits_{x \to 0 } \ \frac{e^{x+3}- \sin x - e^3}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{e^{x+3}- \sin x - e^3}{x} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{e^{3+x} - e^3}{x} - \frac{\sin x}{x} \Bigg] 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ e^3 \Bigg( \frac{e^{x} - 1}{x} \Bigg) - \frac{\sin x}{x} \Bigg] 

\displaystyle = e^3 \times 1 - 1 = e^3 - 1

\\

\displaystyle \text{Question 34: } \lim \limits_{x \to 0 } \ \frac{e^x - x - 1}{2} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{e^x - x - 1}{2} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \infty  \Bigg]   

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{\Big( 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \infty \Big) - x - 1}{2}   \Bigg]   

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \infty }{2}   \Bigg]   

\displaystyle = 0

\\

\displaystyle \text{Question 35: } \lim \limits_{x \to 0 } \ \frac{e^{3x}-e^{2x}}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{e^{3x}-e^{2x}}{x} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[  \frac{3^{3x}-1}{x} - \frac{e^{2x}-1}{x}  \Bigg] 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ 3\Big(  \frac{3^{3x}-1}{x} \Big) - 2 \Big( \frac{e^{2x}-1}{x}  \Big) \Bigg] 

\displaystyle = 3 \times 1 - 2 \times 1 = 1

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\displaystyle \text{Question 36: } \lim \limits_{x \to 0 } \ \frac{e^{\tan x} - 1}{\tan x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{e^{\tan x} - 1}{\tan x} 

\displaystyle \text{If } x \rightarrow 0 , \text{ then } \tan x \rightarrow 0

\displaystyle = \lim \limits_{y \to 0 } \ \frac{e^{y} - 1}{y} 

\displaystyle = 1

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\displaystyle \text{Question 37: } \lim \limits_{x \to 0 } \ \frac{e^{bx}-e^{ax}}{x} \text{ where } 0 < a < b

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{e^{bx}-e^{ax}}{x}

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{e^{bx}-1}{x} - \frac{e^{ax}-1}{x}  \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ b \Bigg( \frac{e^{bx}-1}{x} \Bigg)  - a \Bigg( \frac{e^{ax}-1}{x} \Bigg)  \Bigg]

\displaystyle = b \times 1 - a \times 1 = b - a

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\displaystyle \text{Question 38: } \lim \limits_{x \to 0 } \ \frac{e^{\tan x} - 1}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \ \frac{e^{\tan x} - 1}{x} 

\displaystyle = \lim \limits_{x \to 0 } \Bigg[  \frac{e^{\tan x} -1 }{\tan x} \times \frac{\tan x}{x} \Bigg]

\displaystyle = 1 \times 1

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\displaystyle \text{Question 39: } \lim \limits_{x \to 0 } \ \frac{e^x - e^{\sin x}}{x - \sin x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \Bigg[  \frac{e^x - e^{\sin x}}{x - \sin x}  \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[  \frac{ e^{\sin x} \Big( \frac{e^x}{e^{\sin x}} - 1 \Big)}{x - \sin x}  \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \ e^{\sin x} \Bigg[  \frac{ e^{x-\sin x}-1}{x - \sin x}  \Bigg]

\displaystyle = e^{\sin 0} = 1

\\

\displaystyle \text{Question 40: } \lim \limits_{x \to 0 } \ \frac{3^{2+x} - 9}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \Bigg[ \frac{3^{2+x} - 9}{x} \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{3^2 \cdot 3^x - 3^2}{x} \Bigg]

\displaystyle = 3^2 \lim \limits_{x \to 0 } \Bigg[ \frac{ 3^x - 1}{x} \Bigg]

\displaystyle = 9 \log 3

\\

\displaystyle \text{Question 41: } \lim \limits_{x \to 0 } \ \frac{a^x-a^{-x}}{x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \Bigg[ \frac{a^x-a^{-x}}{x} \Bigg]  

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{a^x-\frac{1}{a^x}}{x} \Bigg]  

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{a^{2x} - 1}{a^x \cdot 2x}  \times 2 \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{a^{2x} - 1}{2x}  \times \frac{2}{a^x}  \Bigg]

\displaystyle = \log a \times \frac{2}{a^0} = 2 \log a

\\

\displaystyle \text{Question 42: } \lim \limits_{x \to 0 } \ \frac{x(e^x-1)}{1 - \cos x} 

Answer:

\displaystyle \lim \limits_{x \to 0 } \Bigg[ \frac{x(e^x-1)}{1 - \cos x} \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{\Big( \frac{e^x-1}{x} \Big)  }{\Big( \frac{1 - \cos x}{x^2} \Big) }  \Bigg]

\displaystyle = \lim \limits_{x \to 0 } \Bigg[ \frac{e^x - 1}{x} \times  \frac{1}{  \frac{2 \sin^2 \frac{x}{2}}{4 \times \frac{x^2}{4}}   } \Bigg]

\displaystyle = 1 \times \frac{2}{1^2} = 2

\\

\displaystyle \text{Question 43: } \lim \limits_{x \to \frac{\pi}{2} } \ \frac{2^{-\cos x} - 1}{x \Big( x - \frac{\pi}{2} \Big)} 

Answer:

\displaystyle \lim \limits_{x \to \frac{\pi}{2} } \Bigg[ \frac{2^{-\cos x} - 1}{x \Big( x - \frac{\pi}{2} \Big)} \Bigg]

\displaystyle = \lim \limits_{x \to \frac{\pi}{2} } \Bigg[ \frac{2^{-\sin ( \frac{\pi}{2} -x )} - 1}{x \Big( x - \frac{\pi}{2} \Big)} \Bigg]

\displaystyle = \lim \limits_{x \to \frac{\pi}{2} } \Bigg[ \frac{2^{\sin ( x - \frac{\pi}{2} )} - 1}{\Big( x - \frac{\pi}{2} \Big) x} \Bigg]

\displaystyle = \lim \limits_{x \to \frac{\pi}{2} } \Bigg[  \frac{2^{\sin ( x - \frac{\pi}{2} )} - 1}{\sin ( x - \frac{\pi}{2} )} \times \frac{\sin ( x - \frac{\pi}{2} )}{( x - \frac{\pi}{2} )} \times \frac{1}{x} \Bigg]

\displaystyle = \log_e 2 \times 1 \times \frac{1}{\frac{\pi}{2}} = \frac{2}{\pi} \log_e 2

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