Question

Evaluate : $limi{t}_{x\to \infty }{\left(\frac{x+2}{x+1}\right)}^{x+3}$

• A. $1$
• B. $e$
• C. $e^2$
• D. $e^3$

Answer 1

The correct option is B

$e$

Explanation for the correct option:

Evaluate the limit

If we substitute the limit it becomes the value ${\left(\frac{\infty }{\infty }\right)}^{\infty }$that does not exist, therefore,

$$\begin{gathered} =limi{t}_{x\to \infty }{\left(\frac{x+2}{x+1}\right)}^{x+3} \\ =limi{t}_{x\to \infty }{\left(\frac{\left(x+1\right)+1}{x+1}\right)}^{\left(x+1\right)+2} \\ =limi{t}_{x\to \infty }{\left(1+\frac{1}{x+1}\right)}^{\left(x+1\right)+2} \\ =limi{t}_{x\to \infty }{\left(1+\frac{1}{x+1}\right)}^{\left(x+1\right)}.{\left(1+\frac{1}{x+1}\right)}^2\left[\because a^{x+y}=a^x.a^y\right] \\ =limi{t}_{x\to \infty }{\left(1+\frac{1}{x+1}\right)}^{\left(x+1\right)}limi{t}_{x\to \infty }{\left(1+\frac{1}{x+1}\right)}^2 \\ =e.{\left(1+\frac{1}{\infty +1}\right)}^2\mathbf{}\left[\because limi{t}_{x\to \infty }{\left(1+\frac{1}{x}\right)}^x=e\right] \\ =e.{\left(1+0\right)}^2\mathbf{}\left[\because \frac{1}{\infty }=0\right]\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ =e \end{gathered}$$

Hence, option (B) is the correct answer