Class 11: Limits – Exercise 29.10

Note:

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

Evaluate the following limits:

$\displaystyle \textbf{Question 1: }\lim \limits_{x\to0}\frac{5^x-1}{\sqrt{4+x}-2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{5^x-1}{\sqrt{4+x}-2}$
$\displaystyle =\lim \limits_{x\to0}\frac{(5^x-1)(\sqrt{4+x}+2)}{(\sqrt{4+x}-2)(\sqrt{4+x}+2)}$
$\displaystyle =\lim \limits_{x\to0}\frac{(5^x-1)(\sqrt{4+x}+2)}{4+x-4}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{5^x-1}{x}\right)(\sqrt{4+x}+2)$
$\displaystyle =\left(\lim \limits_{x\to0}\frac{5^x-1}{x}\right)\left(\lim \limits_{x\to0}(\sqrt{4+x}+2)\right)$
$\displaystyle =\log 5\,(2+2)$
$\displaystyle =4\log 5$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{5^x-1}{\sqrt{4+x}-2}=4\log 5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\lim \limits_{x\to0}\frac{\log(1+x)}{3^x-1}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{\log(1+x)}{3^x-1}$
$\displaystyle =\lim \limits_{x\to0}\frac{\dfrac{\log(1+x)}{x}}{\dfrac{3^x-1}{x}}$
$\displaystyle =\frac{\displaystyle \lim \limits_{x\to0}\frac{\log(1+x)}{x}}{\displaystyle \lim \limits_{x\to0}\frac{3^x-1}{x}}$
$\displaystyle =\frac{1}{\log 3}$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{\log(1+x)}{3^x-1}=\frac{1}{\log 3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\lim \limits_{x\to0}\frac{a^x+a^{-x}-2}{x^2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{a^x+a^{-x}-2}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\frac{\left(a^{\frac{x}{2}}\right)^2+\left(a^{-\frac{x}{2}}\right)^2-2a^{\frac{x}{2}}a^{-\frac{x}{2}}}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\frac{\left(a^{\frac{x}{2}}-a^{-\frac{x}{2}}\right)^2}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\frac{\left(a^{\frac{x}{2}}-\frac{1}{a^{\frac{x}{2}}}\right)^2}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\frac{\left(\frac{a^x-1}{a^{\frac{x}{2}}}\right)^2}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{a^x-1}{x}\right)^2\frac{1}{a^x}$
$\displaystyle =(\log a)^2\cdot\frac{1}{a^0}$
$\displaystyle =(\log a)^2$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{a^x+a^{-x}-2}{x^2}=(\log a)^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\lim \limits_{x\to0}\frac{a^{mx}-1}{b^{nx}-1},\quad n\neq0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{a^{mx}-1}{b^{nx}-1}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{a^{mx}-1}{mx}\right)\left(\frac{nx}{b^{nx}-1}\right)\frac{m}{n}$
$\displaystyle =\left(\lim \limits_{x\to0}\frac{a^{mx}-1}{mx}\right)\left(\lim \limits_{x\to0}\frac{nx}{b^{nx}-1}\right)\frac{m}{n}$
$\displaystyle =(\log a)\left(\frac{1}{\log b}\right)\frac{m}{n}$
$\displaystyle =\frac{m}{n}\frac{\log a}{\log b}$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{a^{mx}-1}{b^{nx}-1}=\frac{m}{n}\frac{\log a}{\log b}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\lim \limits_{x\to0}\frac{a^x+b^x-2}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{a^x+b^x-2}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{(a^x-1)+(b^x-1)}{x}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{a^x-1}{x}+\frac{b^x-1}{x}\right)$
$\displaystyle =\lim \limits_{x\to0}\frac{a^x-1}{x}+\lim \limits_{x\to0}\frac{b^x-1}{x}$
$\displaystyle =\log a+\log b$
$\displaystyle =\log(ab)$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{a^x+b^x-2}{x}=\log(ab).$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\lim \limits_{x\to0}\frac{9^x-2\cdot6^x+4^x}{x^2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{9^x-2\cdot6^x+4^x}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\frac{(3^x)^2-2\cdot3^x\cdot2^x+(2^x)^2}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\frac{(3^x-2^x)^2}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\left[\frac{2^x\left(\left(\frac{3}{2}\right)^x-1\right)}{x}\right]^2$
$\displaystyle =\lim \limits_{x\to0}2^{2x}\left[\frac{\left(\frac{3}{2}\right)^x-1}{x}\right]^2$
$\displaystyle =2^0\left[\log\left(\frac{3}{2}\right)\right]^2$
$\displaystyle =\left[\log\left(\frac{3}{2}\right)\right]^2$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{9^x-2\cdot6^x+4^x}{x^2}=\left[\log\left(\frac{3}{2}\right)\right]^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\lim \limits_{x\to0}\frac{8^x-4^x-2^x+1}{x^2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{8^x-4^x-2^x+1}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\frac{(2^x)^3-(2^x)^2-2^x+1}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\frac{(2^x)^2(2^x-1)-(2^x-1)}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\frac{(2^{2x}-1)(2^x-1)}{x^2}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{2(2^{2x}-1)}{2x}\right)\left(\frac{2^x-1}{x}\right)$
$\displaystyle =2\left(\lim \limits_{x\to0}\frac{2^{2x}-1}{2x}\right)\left(\lim \limits_{x\to0}\frac{2^x-1}{x}\right)$
$\displaystyle =2\log2\cdot\log2$
$\displaystyle =2(\log2)^2$
$\displaystyle =(\log4)(\log2)$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{8^x-4^x-2^x+1}{x^2}=2(\log2)^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\lim \limits_{x\to0}\frac{a^{mx}-b^{nx}}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{a^{mx}-b^{nx}}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{(a^{mx}-1)-(b^{nx}-1)}{x}$
$\displaystyle =\lim \limits_{x\to0}\left[\left(\frac{a^{mx}-1}{mx}\right)m-\left(\frac{b^{nx}-1}{nx}\right)n\right]$
$\displaystyle =m\log a-n\log b$
$\displaystyle =\log a^m-\log b^n$
$\displaystyle =\log\left(\frac{a^m}{b^n}\right)$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{a^{mx}-b^{nx}}{x}=\log\left(\frac{a^m}{b^n}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\lim \limits_{x\to0}\frac{a^x+b^x+c^x-3}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{a^x+b^x+c^x-3}{x}$
$\displaystyle =\lim \limits_{x\to0}\left[\frac{a^x-1}{x}+\frac{b^x-1}{x}+\frac{c^x-1}{x}\right]$
$\displaystyle =\log a+\log b+\log c$
$\displaystyle =\log(abc)$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{a^x+b^x+c^x-3}{x}=\log(abc).$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\lim \limits_{x\to2}\frac{x-2}{\log_a(x-1)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=2,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \text{Let }x=2+h.$
$\displaystyle \text{As }x\to2,\ h\to0.$
$\displaystyle \lim \limits_{x\to2}\frac{x-2}{\log_a(x-1)}$
$\displaystyle =\lim \limits_{h\to0}\frac{h}{\log_a(1+h)}$
$\displaystyle =\lim \limits_{h\to0}\frac{h}{\dfrac{\log(1+h)}{\log a}}$
$\displaystyle =\log a\lim \limits_{h\to0}\frac{h}{\log(1+h)}$
$\displaystyle =\log a\left(\frac{1}{\displaystyle \lim \limits_{h\to0}\frac{\log(1+h)}{h}}\right)$
$\displaystyle =\log a$
$\displaystyle \therefore \lim \limits_{x\to2}\frac{x-2}{\log_a(x-1)}=\log a.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\lim \limits_{x\to0}\frac{5^x+3^x+2^x-3}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{5^x+3^x+2^x-3}{x}$
$\displaystyle =\lim \limits_{x\to0}\left[\frac{5^x-1}{x}+\frac{3^x-1}{x}+\frac{2^x-1}{x}\right]$
$\displaystyle =\log5+\log3+\log2$
$\displaystyle =\log(5\times3\times2)$
$\displaystyle =\log30$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{5^x+3^x+2^x-3}{x}=\log30.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\lim \limits_{x\to\infty}\left(a^{\frac{1}{x}}-1\right)x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }y=\frac{1}{x}.$
$\displaystyle \text{As }x\to\infty,\ y\to0.$
$\displaystyle \lim \limits_{x\to\infty}\left(a^{\frac{1}{x}}-1\right)x$
$\displaystyle =\lim \limits_{y\to0}(a^y-1)\frac{1}{y}$
$\displaystyle =\lim \limits_{y\to0}\frac{a^y-1}{y}$
$\displaystyle =\log a$
$\displaystyle \therefore \lim \limits_{x\to\infty}\left(a^{\frac{1}{x}}-1\right)x=\log a.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\lim \limits_{x\to0}\frac{a^{mx}-b^{nx}}{\sin kx}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{a^{mx}-b^{nx}}{\sin kx}$
$\displaystyle =\lim \limits_{x\to0}\frac{(a^{mx}-1)-(b^{nx}-1)}{\sin kx}$
$\displaystyle =\lim \limits_{x\to0}\frac{m\left(\frac{a^{mx}-1}{mx}\right)-n\left(\frac{b^{nx}-1}{nx}\right)}{k\left(\frac{\sin kx}{kx}\right)}$
$\displaystyle =\frac{m\log a-n\log b}{k}$
$\displaystyle =\frac{1}{k}\left[\log(a^m)-\log(b^n)\right]$
$\displaystyle =\frac{1}{k}\log\left(\frac{a^m}{b^n}\right)$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{a^{mx}-b^{nx}}{\sin kx}=\frac{1}{k}\log\left(\frac{a^m}{b^n}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\lim \limits_{x\to0}\frac{a^x+b^x-c^x-d^x}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{a^x+b^x-c^x-d^x}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{(a^x-1)+(b^x-1)-(c^x-1)-(d^x-1)}{x}$
$\displaystyle =\lim \limits_{x\to0}\left[\frac{a^x-1}{x}+\frac{b^x-1}{x}-\frac{c^x-1}{x}-\frac{d^x-1}{x}\right]$
$\displaystyle =\log a+\log b-\log c-\log d$
$\displaystyle =\log\left(\frac{ab}{cd}\right)$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{a^x+b^x-c^x-d^x}{x}=\log\left(\frac{ab}{cd}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\lim \limits_{x\to0}\frac{e^x-1+\sin x}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{e^x-1+\sin x}{x}$
$\displaystyle =\lim \limits_{x\to0}\left[\frac{e^x-1}{x}+\frac{\sin x}{x}\right]$
$\displaystyle =\lim \limits_{x\to0}\frac{e^x-1}{x}+\lim \limits_{x\to0}\frac{\sin x}{x}$
$\displaystyle =1+1$
$\displaystyle =2$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^x-1+\sin x}{x}=2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\lim \limits_{x\to0}\frac{\sin2x}{e^x-1}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{\sin2x}{e^x-1}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{\sin2x}{2x}\right)\left(\frac{2x}{e^x-1}\right)$
$\displaystyle =2\left(\lim \limits_{x\to0}\frac{\sin2x}{2x}\right)\left(\lim \limits_{x\to0}\frac{x}{e^x-1}\right)$
$\displaystyle =2\times1\times1$
$\displaystyle =2$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{\sin2x}{e^x-1}=2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\lim \limits_{x\to0}\frac{e^{\sin x}-1}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{e^{\sin x}-1}{x}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{e^{\sin x}-1}{\sin x}\right)\left(\frac{\sin x}{x}\right)$
$\displaystyle \text{Let }y=\sin x.$
$\displaystyle \text{As }x\to0,\ y\to0.$
$\displaystyle =\left(\lim \limits_{y\to0}\frac{e^y-1}{y}\right)\left(\lim \limits_{x\to0}\frac{\sin x}{x}\right)$
$\displaystyle =1\times1$
$\displaystyle =1$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^{\sin x}-1}{x}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\lim \limits_{x\to0}\frac{e^{2x}-e^x}{\sin2x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{e^{2x}-e^x}{\sin2x}$
$\displaystyle =\lim \limits_{x\to0}\frac{e^x(e^x-1)}{\sin2x}$
$\displaystyle =\lim \limits_{x\to0}\left[e^x\left(\frac{e^x-1}{x}\right)\left(\frac{2x}{\sin2x}\right)\times\frac{1}{2}\right]$
$\displaystyle =e^0\times1\times1\times\frac{1}{2}$
$\displaystyle =\frac{1}{2}$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^{2x}-e^x}{\sin2x}=\frac{1}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\lim \limits_{x\to a}\frac{\log x-\log a}{x-a},\quad a>0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=a,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to a}\frac{\log x-\log a}{x-a}$
$\displaystyle =\lim \limits_{x\to a}\frac{\log\left(\frac{x}{a}\right)}{a\left(\frac{x}{a}-1\right)}$
$\displaystyle \text{Let }y=\frac{x}{a}-1.$
$\displaystyle \text{As }x\to a,\ y\to0\text{ and }\frac{x}{a}=1+y.$
$\displaystyle =\frac{1}{a}\lim \limits_{y\to0}\frac{\log(1+y)}{y}$
$\displaystyle =\frac{1}{a}\times1$
$\displaystyle =\frac{1}{a}$
$\displaystyle \therefore \lim \limits_{x\to a}\frac{\log x-\log a}{x-a}=\frac{1}{a}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\lim \limits_{x\to0}\frac{\log(a+x)-\log(a-x)}{x},\quad a>0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{\log(a+x)-\log(a-x)}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{\log\left(\frac{a+x}{a-x}\right)}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{\log\left(1+\frac{2x}{a-x}\right)}{x}$
$\displaystyle =\lim \limits_{x\to0}\left[\frac{\log\left(1+\frac{2x}{a-x}\right)}{\frac{2x}{a-x}}\times\frac{2}{a-x}\right]$
$\displaystyle \text{Let }y=\frac{2x}{a-x}.$
$\displaystyle \text{As }x\to0,\ y\to0.$
$\displaystyle =\left(\lim \limits_{y\to0}\frac{\log(1+y)}{y}\right)\left(\lim \limits_{x\to0}\frac{2}{a-x}\right)$
$\displaystyle =1\times\frac{2}{a}$
$\displaystyle =\frac{2}{a}$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{\log(a+x)-\log(a-x)}{x}=\frac{2}{a}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\lim \limits_{x\to0}\frac{\log(2+x)+\log0.5}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{\log(2+x)+\log0.5}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{\log\left[(2+x)\times0.5\right]}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{\log\left(1+\frac{x}{2}\right)}{x}$
$\displaystyle =\frac{1}{2}\lim \limits_{x\to0}\frac{\log\left(1+\frac{x}{2}\right)}{\frac{x}{2}}$
$\displaystyle =\frac{1}{2}\times1$
$\displaystyle =\frac{1}{2}$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{\log(2+x)+\log0.5}{x}=\frac{1}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\lim \limits_{x\to0}\frac{\log(a+x)-\log a}{x},\quad a>0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{\log(a+x)-\log a}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{\log\left(\frac{a+x}{a}\right)}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{\log\left(1+\frac{x}{a}\right)}{x}$
$\displaystyle =\frac{1}{a}\lim \limits_{x\to0}\frac{\log\left(1+\frac{x}{a}\right)}{\frac{x}{a}}$
$\displaystyle =\frac{1}{a}\times1$
$\displaystyle =\frac{1}{a}$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{\log(a+x)-\log a}{x}=\frac{1}{a}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\lim \limits_{x\to0}\frac{\log(3+x)-\log(3-x)}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{\log(3+x)-\log(3-x)}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{\log\left(\frac{3+x}{3-x}\right)}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{\log\left(1+\frac{2x}{3-x}\right)}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{\log\left(1+\frac{2x}{3-x}\right)}{\frac{2x}{3-x}}\times\frac{2}{3-x}$
$\displaystyle =1\times\frac{2}{3}$
$\displaystyle =\frac{2}{3}$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{\log(3+x)-\log(3-x)}{x}=\frac{2}{3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\lim \limits_{x\to0}\frac{8^x-2^x}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{8^x-2^x}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{(8^x-1)-(2^x-1)}{x}$
$\displaystyle =\lim \limits_{x\to0}\left[\frac{8^x-1}{x}-\frac{2^x-1}{x}\right]$
$\displaystyle =\log8-\log2$
$\displaystyle =\log\left(\frac{8}{2}\right)$
$\displaystyle =\log4$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{8^x-2^x}{x}=\log4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\lim \limits_{x\to0}\frac{x(2^x-1)}{1-\cos x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{x(2^x-1)}{1-\cos x}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{2^x-1}{x}\right)\left(\frac{x^2}{1-\cos x}\right)$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{2^x-1}{x}\right)\left(\frac{x^2}{2\sin^2\frac{x}{2}}\right)$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{2^x-1}{x}\right)\left[\frac{2}{\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2}\right]$
$\displaystyle =\log2\times\frac{2}{1^2}$
$\displaystyle =2\log2$
$\displaystyle =\log4$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{x(2^x-1)}{1-\cos x}=\log4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\lim \limits_{x\to0}\frac{\sqrt{1+x}-1}{\log(1+x)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{\sqrt{1+x}-1}{\log(1+x)}$
$\displaystyle =\lim \limits_{x\to0}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{\log(1+x)(\sqrt{1+x}+1)}$
$\displaystyle =\lim \limits_{x\to0}\frac{x}{\log(1+x)(\sqrt{1+x}+1)}$
$\displaystyle =\left(\lim \limits_{x\to0}\frac{x}{\log(1+x)}\right)\left(\lim \limits_{x\to0}\frac{1}{\sqrt{1+x}+1}\right)$
$\displaystyle =\frac{1}{\displaystyle \lim \limits_{x\to0}\frac{\log(1+x)}{x}}\times\frac{1}{2}$
$\displaystyle =1\times\frac{1}{2}$
$\displaystyle =\frac{1}{2}$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{\sqrt{1+x}-1}{\log(1+x)}=\frac{1}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\lim \limits_{x\to0}\frac{\log|1+x^3|}{\sin^3x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \text{Since }1+x^3>0\text{ for }x\text{ sufficiently close to }0,\ |1+x^3|=1+x^3.$
$\displaystyle \lim \limits_{x\to0}\frac{\log|1+x^3|}{\sin^3x}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{\log(1+x^3)}{x^3}\right)\left(\frac{x^3}{\sin^3x}\right)$
$\displaystyle =\left(\lim \limits_{x\to0}\frac{\log(1+x^3)}{x^3}\right)\left(\lim \limits_{x\to0}\frac{x}{\sin x}\right)^3$
$\displaystyle =1\times1^3$
$\displaystyle =1$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{\log|1+x^3|}{\sin^3x}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\lim \limits_{x\to\frac{\pi}{2}}\frac{a^{\cot x}-a^{\cos x}}{\cot x-\cos x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=\frac{\pi}{2},\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to\frac{\pi}{2}}\frac{a^{\cot x}-a^{\cos x}}{\cot x-\cos x}$
$\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}a^{\cos x}\left(\frac{\dfrac{a^{\cot x}}{a^{\cos x}}-1}{\cot x-\cos x}\right)$
$\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}a^{\cos x}\left(\frac{a^{\cot x-\cos x}-1}{\cot x-\cos x}\right)$
$\displaystyle \text{Let }y=\cot x-\cos x.$
$\displaystyle \text{As }x\to\frac{\pi}{2},\ y\to0.$
$\displaystyle =a^0\left(\lim \limits_{y\to0}\frac{a^y-1}{y}\right)$
$\displaystyle =1\times\log a$
$\displaystyle =\log a$
$\displaystyle \therefore \lim \limits_{x\to\frac{\pi}{2}}\frac{a^{\cot x}-a^{\cos x}}{\cot x-\cos x}=\log a.$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\lim \limits_{x\to0}\frac{e^x-1}{\sqrt{1-\cos x}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \text{Using }1-\cos x=2\sin^2\frac{x}{2},$
$\displaystyle \sqrt{1-\cos x}=\sqrt{2}\left|\sin\frac{x}{2}\right|.$
$\displaystyle \therefore \frac{e^x-1}{\sqrt{1-\cos x}}=\frac{e^x-1}{\sqrt{2}\left|\sin\frac{x}{2}\right|}$
$\displaystyle =\frac{e^x-1}{x}\times\frac{x}{2\left|\sin\frac{x}{2}\right|}\times\sqrt{2}.$
$\displaystyle \text{Left-hand limit:}$
$\displaystyle \lim \limits_{x\to0^-}\frac{e^x-1}{\sqrt{1-\cos x}}$
$\displaystyle =\sqrt{2}\left(\lim \limits_{x\to0^-}\frac{e^x-1}{x}\right)\left(\lim \limits_{x\to0^-}\frac{x}{2\left|\sin\frac{x}{2}\right|}\right)$
$\displaystyle =\sqrt{2}\times1\times(-1)$
$\displaystyle =-\sqrt{2}.$
$\displaystyle \text{Right-hand limit:}$
$\displaystyle \lim \limits_{x\to0^+}\frac{e^x-1}{\sqrt{1-\cos x}}$
$\displaystyle =\sqrt{2}\left(\lim \limits_{x\to0^+}\frac{e^x-1}{x}\right)\left(\lim \limits_{x\to0^+}\frac{x}{2\left|\sin\frac{x}{2}\right|}\right)$
$\displaystyle =\sqrt{2}\times1\times1$
$\displaystyle =\sqrt{2}.$
$\displaystyle \text{Since the left-hand limit is not equal to the right-hand limit,}$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^x-1}{\sqrt{1-\cos x}}\text{ does not exist.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\lim \limits_{x\to5}\frac{e^x-e^5}{x-5}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=5,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to5}\frac{e^x-e^5}{x-5}$
$\displaystyle =\lim \limits_{x\to5}e^5\left(\frac{e^{x-5}-1}{x-5}\right)$
$\displaystyle \text{Let }y=x-5.$
$\displaystyle \text{As }x\to5,\ y\to0.$
$\displaystyle =e^5\lim \limits_{y\to0}\frac{e^y-1}{y}$
$\displaystyle =e^5\times1$
$\displaystyle =e^5$
$\displaystyle \therefore \lim \limits_{x\to5}\frac{e^x-e^5}{x-5}=e^5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\lim \limits_{x\to0}\frac{e^{x+2}-e^2}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{e^{x+2}-e^2}{x}$
$\displaystyle =\lim \limits_{x\to0}e^2\left(\frac{e^x-1}{x}\right)$
$\displaystyle =e^2\lim \limits_{x\to0}\frac{e^x-1}{x}$
$\displaystyle =e^2\times1$
$\displaystyle =e^2$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^{x+2}-e^2}{x}=e^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\lim \limits_{x\to\frac{\pi}{2}}\frac{e^{\cos x}-1}{\cos x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=\frac{\pi}{2},\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \text{Let }y=\cos x.$
$\displaystyle \text{As }x\to\frac{\pi}{2},\ y\to0.$
$\displaystyle \lim \limits_{x\to\frac{\pi}{2}}\frac{e^{\cos x}-1}{\cos x}$
$\displaystyle =\lim \limits_{y\to0}\frac{e^y-1}{y}$
$\displaystyle =1$
$\displaystyle \therefore \lim \limits_{x\to\frac{\pi}{2}}\frac{e^{\cos x}-1}{\cos x}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\lim \limits_{x\to0}\frac{e^{x+3}-\sin x-e^3}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{e^{x+3}-\sin x-e^3}{x}$
$\displaystyle =\lim \limits_{x\to0}\left[\frac{e^{x+3}-e^3}{x}-\frac{\sin x}{x}\right]$
$\displaystyle =\lim \limits_{x\to0}\left[e^3\left(\frac{e^x-1}{x}\right)-\frac{\sin x}{x}\right]$
$\displaystyle =e^3\times1-1$
$\displaystyle =e^3-1$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^{x+3}-\sin x-e^3}{x}=e^3-1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\lim \limits_{x\to0}\frac{e^x-x-1}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{By direct substitution,}$
$\displaystyle \lim \limits_{x\to0}\frac{e^x-x-1}{2}$
$\displaystyle =\frac{e^0-0-1}{2}$
$\displaystyle =\frac{1-1}{2}$
$\displaystyle =0$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^x-x-1}{2}=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\lim \limits_{x\to0}\frac{e^{3x}-e^{2x}}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{e^{3x}-e^{2x}}{x}$
$\displaystyle =\lim \limits_{x\to0}\left[\frac{e^{3x}-1}{x}-\frac{e^{2x}-1}{x}\right]$
$\displaystyle =\lim \limits_{x\to0}\left[3\left(\frac{e^{3x}-1}{3x}\right)-2\left(\frac{e^{2x}-1}{2x}\right)\right]$
$\displaystyle =3\times1-2\times1$
$\displaystyle =1$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^{3x}-e^{2x}}{x}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\lim \limits_{x\to0}\frac{e^{\tan x}-1}{\tan x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \text{Let }y=\tan x.$
$\displaystyle \text{As }x\to0,\ y\to0.$
$\displaystyle \lim \limits_{x\to0}\frac{e^{\tan x}-1}{\tan x}=\lim \limits_{y\to0}\frac{e^y-1}{y}$
$\displaystyle =1$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^{\tan x}-1}{\tan x}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: }\lim \limits_{x\to0}\frac{e^{bx}-e^{ax}}{x},\ \text{where }0<a<b$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{e^{bx}-e^{ax}}{x}$
$\displaystyle =\lim \limits_{x\to0}\left[\frac{e^{bx}-1}{x}-\frac{e^{ax}-1}{x}\right]$
$\displaystyle =\lim \limits_{x\to0}\left[b\left(\frac{e^{bx}-1}{bx}\right)-a\left(\frac{e^{ax}-1}{ax}\right)\right]$
$\displaystyle =b\times1-a\times1$
$\displaystyle =b-a$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^{bx}-e^{ax}}{x}=b-a.$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: }\lim \limits_{x\to0}\frac{e^{\tan x}-1}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{e^{\tan x}-1}{x}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{e^{\tan x}-1}{\tan x}\right)\left(\frac{\tan x}{x}\right)$
$\displaystyle =1\times1$
$\displaystyle =1$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^{\tan x}-1}{x}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 39: }\lim \limits_{x\to0}\frac{e^x-e^{\sin x}}{x-\sin x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{e^x-e^{\sin x}}{x-\sin x}$
$\displaystyle =\lim \limits_{x\to0}e^{\sin x}\left(\frac{e^{x-\sin x}-1}{x-\sin x}\right)$
$\displaystyle \text{Let }y=x-\sin x.$
$\displaystyle \text{As }x\to0,\ y\to0.$
$\displaystyle =\left(\lim \limits_{x\to0}e^{\sin x}\right)\left(\lim \limits_{y\to0}\frac{e^y-1}{y}\right)$
$\displaystyle =e^0\times1$
$\displaystyle =1$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{e^x-e^{\sin x}}{x-\sin x}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 40: }\lim \limits_{x\to0}\frac{3^{2+x}-9}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{3^{2+x}-9}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{3^2\cdot3^x-3^2}{x}$
$\displaystyle =9\lim \limits_{x\to0}\frac{3^x-1}{x}$
$\displaystyle =9\log_e3$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{3^{2+x}-9}{x}=9\log_e3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 41: }\lim \limits_{x\to0}\frac{a^x-a^{-x}}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{a^x-a^{-x}}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{a^x-\frac{1}{a^x}}{x}$
$\displaystyle =\lim \limits_{x\to0}\frac{a^{2x}-1}{a^x x}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{a^{2x}-1}{2x}\right)\left(\frac{2}{a^x}\right)$
$\displaystyle =\log_e a\times\frac{2}{a^0}$
$\displaystyle =2\log_e a$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{a^x-a^{-x}}{x}=2\log_e a.$
$\displaystyle \\$

$\displaystyle \textbf{Question 42: }\lim \limits_{x\to0}\frac{x(e^x-1)}{1-\cos x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=0,\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \lim \limits_{x\to0}\frac{x(e^x-1)}{1-\cos x}$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{e^x-1}{x}\right)\left(\frac{x^2}{1-\cos x}\right)$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{e^x-1}{x}\right)\left(\frac{x^2}{2\sin^2\frac{x}{2}}\right)$
$\displaystyle =\lim \limits_{x\to0}\left(\frac{e^x-1}{x}\right)\left[\frac{2}{\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2}\right]$
$\displaystyle =1\times\frac{2}{1^2}$
$\displaystyle =2$
$\displaystyle \therefore \lim \limits_{x\to0}\frac{x(e^x-1)}{1-\cos x}=2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 43: }\lim \limits_{x\to\frac{\pi}{2}}\frac{2^{-\cos x}-1}{x\left(x-\frac{\pi}{2}\right)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=\frac{\pi}{2},\text{ the expression assumes the form }\frac{0}{0}.$
$\displaystyle \text{Since }-\cos x=\sin\left(x-\frac{\pi}{2}\right),$
$\displaystyle \lim \limits_{x\to\frac{\pi}{2}}\frac{2^{-\cos x}-1}{x\left(x-\frac{\pi}{2}\right)}$
$\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}\frac{2^{\sin\left(x-\frac{\pi}{2}\right)}-1}{x\left(x-\frac{\pi}{2}\right)}$
$\displaystyle =\lim \limits_{x\to\frac{\pi}{2}}\left[\frac{2^{\sin\left(x-\frac{\pi}{2}\right)}-1}{\sin\left(x-\frac{\pi}{2}\right)}\right]$
$\displaystyle \times\left[\frac{\sin\left(x-\frac{\pi}{2}\right)}{x-\frac{\pi}{2}}\right]\times\frac{1}{x}$
$\displaystyle =\log_e2\times1\times\frac{1}{\frac{\pi}{2}}$
$\displaystyle =\frac{2\log_e2}{\pi}$
$\displaystyle \therefore \lim \limits_{x\to\frac{\pi}{2}}\frac{2^{-\cos x}-1}{x\left(x-\frac{\pi}{2}\right)}=\frac{2\log_e2}{\pi}.$
$\displaystyle \\$