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

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

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$e^5$

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(C) We use the standard limit formula: $\lim _{x \rightarrow \infty} (1 + \frac{a}{x+b})^{x+c} = e^a$. Given the expression: $\lim _{x \rightarrow \infty} (\frac{x+6}{x+1})^{x+4}$. Rewrite the base: $\frac{x+6}{x+1} = \frac{x+1+5}{x+1} = 1 + \frac{5}{x+1}$. So,the limit becomes $\lim _{x \rightarrow \infty} (1 + \frac{5}{x+1})^{x+4}$. Using the property $\lim _{x \rightarrow \infty} (1 + \frac{k}{f(x)})^{g(x)} = e^{\lim _{x \rightarrow \infty} k \cdot \frac{g(x)}{f(x)}}$: $= e^{\lim _{x \rightarrow \infty} 5 \cdot \frac{x+4}{x+1}}$. $= e^{5 \cdot \lim _{x \rightarrow \infty} \frac{1 + 4/x}{1 + 1/x}} = e^{5 \cdot 1} = e^5$.

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

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