Q103 / 160 · AP EAMCET 2021 (24 Aug Shift 1) · Mathematics

$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to

  • A
    $e^4$
  • B
    $e^6$
  • Correct answer
    C
    $e^5$
  • D
    $e$

Solution

$$ \begin{aligned} & \text { } \lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4} \\ & \lim _{x \rightarrow \infty}\left(\frac{1+\frac{6}{x}}{1+\frac{1}{x}}\right)^{x+4} \\ & x \rightarrow \infty, 1+\frac{6}{x} \rightarrow 1,1+\frac{1}{x} \rightarrow 1 \\ & \qquad \frac{1+\frac{6}{x}}{1+\frac{1}{x}} \rightarrow 1 \\ & \text { and } x+4 \rightarrow \infty \end{aligned} $$ $\therefore$ It is an indeterminate form of $1^{\infty}$. Now, $\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}=e^{\lim _{x \rightarrow-\left(\frac{x+6}{x+1}-1\right) \cdot x+4}}$ $$ \begin{aligned} & =e^{\lim _{x \rightarrow-\infty}\left(\frac{x+6-x-1}{x+1}\right) \cdot x+4} \\ & =e^{\lim _{x \rightarrow-\infty} \frac{5(x+4)}{x+1}}=e^{\lim _{x \rightarrow-\infty} \frac{x\left(1+\frac{4}{x}\right)}{x\left(1+\frac{1}{x}\right)}} \\ & =e^{5 \frac{(1+0)}{(1+0)}} \\ & {\left[\therefore x \rightarrow \infty, \frac{1}{x} \rightarrow 0\right]} \\ & =e^5 \\ & \end{aligned} $$

Hence, the correct answer is option: 'C'.

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