Question

$\underset{x\to \infty }{\lim }{\left(1+\frac{4}{x-1}\right)}^{x+3}=$

• A. $e^4$
• B. $e^2$
• C. $e^3$
• D. $e$

Answer 1

The correct option is A

$e^4$

Explanation for the correct option:

Finding the value of $\underset{x\to \infty }{\lim }{\left(1+\frac{4}{x-1}\right)}^{x+3}$:

The given value is,

$I=\underset{x\to \infty }{\lim }{\left(1+\frac{4}{x-1}\right)}^{x+3}$

Rewriting the above equation as,

$$\begin{gathered} I=\underset{x\to \infty }{\lim }{\left(1+\frac{4}{x-1}\right)}^{\frac{x-1}{4}\frac{4\left(x+3\right)}{x-1}} \\ \Rightarrow I=e^{\underset{x\to \infty }{\lim }\frac{4\left(x+3\right)}{x-1}}\left[\because 1^{\infty }\mathrm{form}\right] \\ \Rightarrow I=e^{4\underset{x\to \infty }{\lim }\frac{x\left(1+{\displaystyle \frac{3}{x}}\right)}{x\left(1-{\displaystyle \frac{1}{x}}\right)}} \\ \Rightarrow I=e^{4\underset{x\to \infty }{\lim }\frac{\left(1+{\displaystyle \frac{3}{x}}\right)}{\left(1-{\displaystyle \frac{1}{x}}\right)}} \end{gathered}$$

At,

$$\begin{gathered} x\to \infty \\ \frac{1}{x}\to \frac{1}{\infty }\to 0 \end{gathered}$$

Then,

$$\begin{gathered} I=e^{4\frac{\left(1+{\displaystyle 0}\right)}{\left(1-{\displaystyle 0}\right)}} \\ \Rightarrow I=e^{4\times 1} \\ \Rightarrow I=e^4 \end{gathered}$$

Hence, the correct answer is option (A).