Question

$\underset{x\to 0}{lim}$ $(\frac{(1+x{)}^2}{e^x}{)}^{\frac{4}{\sin x}}$ is:

• A. $e^2$
• B. $e^4$
• C. $e^8$
• D. $e^{-8}$

Answer 1

The correct option is B $e^4$

Now

$\underset{x\to 0}{lim}{\left(\frac{(1+x{)}^2}{e^x}\right)}^{\frac{4}{\sin x}}$

$=\underset{x\to 0}{lim}\frac{{\left({\left\{\right(1+x{)}^{\frac{1}{x}}\}}^{\frac{x}{\sin x}}\right)}^8}{e^{\frac{4x}{\sin x}}}$

We have $\underset{x\to 0}{lim}\frac{\sin x}{x}=1$ and $\underset{x\to 0}{lim}(1+x{)}^{\frac{1}{x}}=e$.

As both the limits of the numerator and denominator exists,and the limit of the numerator is non-vanishing, so we can write the above limits as

$=\frac{\underset{x\to 0}{lim}{\left({\left\{\right(1+x{)}^{\frac{1}{x}}\}}^{\underset{x\to 0}{lim}\frac{x}{\sin x}}\right)}^8}{\underset{x\to 0}{lim}{e}^{\frac{4x}{\sin x}}}$ [Using division property of limits]

$=\frac{{\left(\underset{x\to 0}{lim}{\left\{\right(1+x{)}^{\frac{1}{x}}\}}^{\left(\underset{x\to 0}{lim}\frac{x}{\sin x}\right)}\right)}^8}{e^{\left(\underset{x\to 0}{lim}\frac{4x}{\sin x}\right)}}$ [Using limit property]

$=\frac{(e^1{)}^8}{e^{\frac{4}{1}}}=e^2$