Question

Arrange the following limits in the ascending order of their values
$\left(1\right)\underset{x\to \infty }{lim}{\left(\frac{1+x}{2+x}\right)}^{x+2}$
$\left(2\right)\underset{x\to 0}{lim}{(1+2x)}^{3/x}$
$\left(3\right)\underset{\theta \to 0}{lim}\left(\frac{\sin \theta }{2\theta }\right)$
$\left(4\right)\underset{x\to 0}{lim}\frac{\ln (1+x)}{x}$

• 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 B $1,3,4,2$

Using standard formulae
$\left(1\right)e^{\underset{x\to \infty }{lim}(2+x)\left(\frac{1+x-2-x}{2+x}\right)}=e^{-1}=\frac{1}{e}$
$\left(2\right)e^{\underset{x\to 0}{lim}\frac{3}{x}(1+2x-1)}=e^6$
$\left(3\right)\underset{x\to 0}{lim}\frac{\sin \theta }{\theta }.\frac{1}{2}=\frac{1}{2}$
$\left(4\right)\underset{x\to 0}{lim}\frac{\ln (1+x)}{x}=1$