lim _{x rightarrow infty}left(frac{x+8}{x+1}r | vedclass

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(D) We know that $\lim _{x \rightarrow \infty} (1 + \frac{a}{x})^x = e^a$. Given expression is $L = \lim _{x \rightarrow \infty} (\frac{x+8}{x+1})^{x+

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$e^7$

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(D) We know that $\lim _{x \rightarrow \infty} (1 + \frac{a}{x})^x = e^a$. Given expression is $L = \lim _{x \rightarrow \infty} (\frac{x+8}{x+1})^{x+5}$. Rewrite the base: $\frac{x+8}{x+1} = \frac{x+1+7}{x+1} = 1 + \frac{7}{x+1}$. So,$L = \lim _{x \rightarrow \infty} (1 + \frac{7}{x+1})^{x+5}$. Let $u = x+1$,then as $x \rightarrow \infty$,$u \rightarrow \infty$ and $x = u-1$. $L = \lim _{u \rightarrow \infty} (1 + \frac{7}{u})^{u-1+5} = \lim _{u \rightarrow \infty} (1 + \frac{7}{u})^{u+4}$. $L = \lim _{u \rightarrow \infty} (1 + \frac{7}{u})^u \cdot \lim _{u \rightarrow \infty} (1 + \frac{7}{u})^4$. $L = e^7 \cdot (1 + 0)^4 = e^7 \cdot 1 = e^7$.

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

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