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

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(C) We have the limit $L = \lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$.Since this is an indeterminate form of $1^{\infty}$,we use

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$e^5$

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(C) We have the limit $L = \lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$.Since this is an indeterminate form of $1^{\infty}$,we use the formula $\lim _{x \rightarrow a} f(x)^{g(x)} = e^{\lim _{x \rightarrow a} (f(x)-1)g(x)}$.$L = e^{\lim _{x \rightarrow \infty} \left(\frac{x+6}{x+1} - 1\right)(x+4)}$$L = e^{\lim _{x \rightarrow \infty} \left(\frac{x+6-x-1}{x+1}\right)(x+4)}$$L = e^{\lim _{x \rightarrow \infty} \frac{5(x+4)}{x+1}}$$L = e^{\lim _{x \rightarrow \infty} \frac{5x+20}{x+1}}$Dividing numerator and denominator by $x$:$L = e^{\lim _{x \rightarrow \infty} \frac{5 + 20/x}{1 + 1/x}}$As $x \rightarrow \infty$,$1/x \rightarrow 0$,so:$L = e^{\frac{5+0}{1+0}} = e^5$.

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

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