Question

Arrange the following limits in the ascending order :
(1) ${lim}_{x\to \infty }{\left(\frac{1+x}{2+x}\right)}^{x+2}$
(2) ${lim}_{x\to 0}(1+2x{)}^{3/x}$
(3) ${lim}_{\theta \to 0}\frac{\sin \theta }{2\theta }$
(4) ${lim}_{x\to 0}\frac{{\log }_e(1+x)}{x}$

• A. $1,2,3,4$
• B. $1,3,4,2$
• C. $1,4,3,2$
• D. $3,4,1,2$

Answer 1

The correct option is D $3,4,1,2$

(i) $\underset{x\to \infty }{lim}{\left(\frac{1+x}{2+x}\right)}^{x+2}$

$=$ ${\left(\frac{\frac{1}{x}+1}{\frac{2}{x}+1}\right)}^{x+2}$

$=\underset{x\to \infty }{lim}{e}^{ln{\left(\frac{1+x}{2+x}\right)}^{2+x}}$

$=e^{\underset{x\to \infty }{lt}(2+x)ln\left(\frac{1+x}{2+x}\right)}$

$=e^{\underset{x\to \infty }{lt}\frac{ln(1+x)-ln(2+x)}{(1/2+x)}}$

$=e^{\underset{x\to \infty }{lt}\frac{\frac{1}{1+x}-\frac{1}{2+x}}{\frac{-1}{{(2+x)}^2}}}$

$=e^{\underset{x\to \infty }{lt}\frac{{(2+x)}^2[(2+x)-(1+x)]}{(2+x)-(1+x)(2+x)(1+x)}}$

$=e^{\underset{x\to \infty }{lt}\frac{(2+x)}{1+x}}$

$=e^1$

(ii) $\underset{x\to 0}{lim}{(1+2x)}^{3/x}$

$=e^{\underset{x\to 0}{lt}\left(2x\right)3/x}$

$=e^6$

(iii) $\underset{\theta \to 0}{lt}\frac{\sin \theta }{2\theta }=\underset{\theta \to 0}{lt}\frac{\cos \theta }{2}=\frac{1}{2}$

(iv) $\underset{x\to 0}{lt}\frac{{log}_e(1+x)}{x}.\underset{x\to 0}{lt}\frac{1}{1+x}=1$

$\therefore$ $3<4<1<2$.