Note:

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

Evaluate the following limits:

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

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

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

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

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

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

$\\$

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

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

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

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

$\\$

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

$\\$

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

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

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

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

$\displaystyle =\log a + \log b$

$\displaystyle = \log (ab)$

$\\$

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

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

$\displaystyle = \log 4 \times \log 2$

$\\$

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

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

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

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

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

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

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

$\\$

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

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

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

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

$\displaystyle = \log (abc)$

$\\$

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

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

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

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

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

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

$\\$

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

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

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

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

$\displaystyle = \log 30$

$\\$

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

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

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

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

$\displaystyle = \log a$

$\\$

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

$\\$

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

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

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

$\displaystyle = 1 + 1 = 2$

$\\$

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

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

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

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

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

$\\$

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

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

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

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

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

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

$\displaystyle = 1 \times 1 = 1$

$\\$

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

$\\$

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

$\\$

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

$\\$

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

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

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

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

$\\$

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

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

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

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

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

$\\$

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

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

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

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

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

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

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.

$\\$

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

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

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

$\displaystyle = e^5$

$\\$

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

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

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

$\displaystyle = e^2$

$\\$

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

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

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

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

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

$\displaystyle = 1$

$\\$

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

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

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

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

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

$\\$

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

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

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

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

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

$\displaystyle = 0$

$\\$

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

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

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

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

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

$\\$

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

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

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

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

$\displaystyle = 1$

$\\$

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

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

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

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

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

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

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

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

$\displaystyle = 1 \times 1$

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

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

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

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

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

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

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

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

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

$\displaystyle = 9 \log 3$

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

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

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

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

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

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

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

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

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

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

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

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

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