Question

For $x$ belongs to $R$, Find$\underset{x\mapsto \infty }{lim}{\left[\frac{(x-3)}{(x+2)}\right]}^x=?$

• A. $e$
• B. $e^{-1}$
• C. $e^{-5}$
• D. $e^5$

Answer 1

The correct option is C

$e^{-5}$

The explanation for the correct option:

Step1. Construction :

We have $\underset{x\mapsto \infty }{lim}{\left[\frac{(x-3)}{(x+2)}\right]}^x$

$$\begin{gathered} =\underset{x\mapsto \infty }{lim}{\left[\frac{(x+2-3-2)}{(x+2)}\right]}^x \\ =\underset{x\mapsto \infty }{lim}{\left[1-\frac{5}{(x+2)}\right]}^x \end{gathered}$$

This is the form of ${\left(1\right)}^{\infty }$and the formula for this

$\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{f}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{e}}^{\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\left[f\left(x\right)-1\right]}$

Step2. Find the limit value :

So, $\underset{x\mapsto \infty }{lim}{\left[1-\frac{5}{(x+2)}\right]}^x=e^{\underset{x\mapsto \infty }{lim}x\left[1-\frac{5}{(x+2)}-1\right]}$

$$\begin{gathered} =e^{\underset{x\mapsto \infty }{lim}x\left[\frac{-5}{(x+2)}\right]} \\ =e^{-5\underset{x\mapsto \infty }{lim}x\left(\frac{1}{(x+2)}\right)} \\ =e^{-5\underset{x\mapsto \infty }{lim}\left(\frac{1}{(1+{\displaystyle \frac{2}{x}})}\right)} \\ =e^{-5} \end{gathered}$$

Hence the correct option is (C).