Question

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

• A. $\sqrt{e}$
• B. $1$
• C. $0$
• D. Does not exist

Answer 1

The correct option is B

$1$

Explanation for correct option.

Step1. Simplifying the integration

Given limit, $\underset{x\to 0}{\lim }\frac{x(e^{{\displaystyle \frac{\sqrt{1+x^2+x^4}-1}{x}}}-1)}{\sqrt{1+x^2+x^4}-1}$

Divide Numerator & denominator with ‘X’.

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{{\displaystyle \frac{x(e^{{\displaystyle \frac{\sqrt{1+x^2+x^4}-1}{x}}}-1)}{x}}}{{\displaystyle \frac{\sqrt{1+x^2+x^4}-1}{x}}} \\ \underset{x\to 0}{\lim }\frac{(e^{\frac{\sqrt{1+x^2+x^4}-1}{x}}-1)}{\frac{\sqrt{1+x^2+x^4}-1}{x}} \end{gathered}$$

Step2. Finding value of limit.

We know that,

$\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$

$\begin{array}{rcl}& & \underset{x\to 0}{\lim }\frac{(e^{\frac{\sqrt{1+x^2+x^4}-1}{x}}-1)}{\frac{\sqrt{1+x^2+x^4}-1}{x}}\\ & =& 1\end{array}$

Hence the correct option is $\mathbf{\left(}\mathit{B}\mathbf{\right)}$.