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MHTCET 22 Apr 2025 Shift 1 Paper

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Question : 134 of 150

Marks: +1, -0

\lim _{x \rightarrow \infty}\left(\frac{x+8}{x+1}\right)^{x+5}=\ldots

$e^{4}$
$e^{5}$
$e^{11}$
$e^{7}$

Solution:

Concept:

This is a

1^\infty

form limit, solved using the standard result

\lim_{x\to\infty}\left(1+\frac{k}{x}\right)^x=e^k

.

Explanation:

Rewrite the base inside the limit:

\frac{x+8}{x+1}=1+\frac{7}{x+1}

So the given expression becomes:

\lim_{x \to \infty}\left(1+\frac{7}{x+1}\right)^{x+5}

Here,

\frac{7}{x+1}\to0

and the exponent

x+5\to\infty

, so it is a

1^\infty

indeterminate form.

Using the standard form, the limit equals

e^{\lim_{x\to\infty} \frac{7}{x+1}(x+5)}

.

Now compute the exponent limit:

\lim_{x\to\infty}\frac{7(x+5)}{x+1}=7

Therefore, the required limit is

e^7

.

Answer:

e^7

, which is option D.

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