Question

$\underset{x\to 0}{\lim }{\left(\frac{1+\tan x}{1+\sin x}\right)}^x=$

• A. $\frac{1}{e}$
• B. $1$
• C. $e$
• D. $e^2$
• E. None of these

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Expanding the given equation and applying the limits:

$\begin{array}{rcl}\underset{x\to 0}{\lim }{\left(\frac{1+\tan x}{1+\sin x}\right)}^2& =& \underset{x\to 0}{\lim }{\left(\frac{1+\tan x}{1+\sin x}\right)}^x\mathbf{[}\mathbf{\because }{\left(\frac{1+tanx}{1+sinx}\right)}^{\mathbf{x}}\mathbf{=}{\mathbf{1}}^{\mathbf{\infty }}\mathbf{f}\mathbf{o}\mathbf{r}\mathbf{m}\mathbf{]}\\ & =& e^{\underset{x\to 0}{\lim }x\left(\frac{1+\tan x}{1+\sin x}-1\right)}\\ & =& e^{\underset{x\to 0}{\lim }x\left(\frac{1+\tan x-1-\sin x}{1+\sin x}\right)}\\ & =& e^{\underset{x\to 0}{\lim }x\left(\frac{\tan x-\sin x}{1+\sin x}\right)}\end{array}$

Note that this is the determinante form.

Applying the limits,

$\begin{array}{rcl}\underset{x\to 0}{\lim }{\left(\frac{1+\tan x}{1+\sin x}\right)}^2& =& e^{0\left(\frac{\tan 0+\sin 0}{1+\sin 0}\right)}\\ & =& e^0\\ & =& 1\end{array}$

Thus, $\underset{x\to 0}{\lim }{\left(\frac{1+\tan x}{1+\sin x}\right)}^x=1$

Therefore, the correct answer is option (B).