lim _{x rightarrow infty}left(frac{x+6}{x+1}r | vedclass

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(C) We know that the standard limit $\lim _{x \rightarrow \infty} (1 + \frac{a}{x+b})^{x+c} = e^a$. Given expression is $\lim _{x \rightarrow \infty}

Vedclass · Accepted Answer

$e^5$

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(C) We know that the standard limit $\lim _{x \rightarrow \infty} (1 + \frac{a}{x+b})^{x+c} = e^a$. Given expression is $\lim _{x \rightarrow \infty} (\frac{x+6}{x+1})^{x+4}$. This can be rewritten as $\lim _{x \rightarrow \infty} (1 + \frac{x+6}{x+1} - 1)^{x+4} = \lim _{x \rightarrow \infty} (1 + \frac{5}{x+1})^{x+4}$. Using the standard limit formula $\lim _{x \rightarrow \infty} (1 + \frac{k}{f(x)})^{g(x)} = e^{\lim _{x \rightarrow \infty} k \cdot \frac{g(x)}{f(x)}}$,we get: $e^{\lim _{x \rightarrow \infty} 5 \cdot \frac{x+4}{x+1}} = e^{5 \cdot \lim _{x \rightarrow \infty} \frac{1+4/x}{1+1/x}} = e^{5 \cdot 1} = e^5$.

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

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