lim _{x rightarrow infty} left(frac{x+a}{x+b} | vedclass

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(A) We evaluate the limit $\lim _{x \rightarrow \infty} \left(\frac{x+a}{x+b}\right)^{x}$.This is of the form $1^{\infty}$.We can rewrite the expressi

Vedclass · Accepted Answer

$e^{a-b}$

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(A) We evaluate the limit $\lim _{x \rightarrow \infty} \left(\frac{x+a}{x+b}\right)^{x}$.This is of the form $1^{\infty}$.We can rewrite the expression as:$\lim _{x \rightarrow \infty} \left(\frac{x+b+a-b}{x+b}\right)^{x} = \lim _{x \rightarrow \infty} \left(1 + \frac{a-b}{x+b}\right)^{x}$.Using the standard limit formula $\lim _{x \rightarrow \infty} (1 + \frac{k}{x})^x = e^k$,we have:$\lim _{x \rightarrow \infty} \left(1 + \frac{a-b}{x+b}\right)^{x} = \lim _{x \rightarrow \infty} \left[\left(1 + \frac{a-b}{x+b}\right)^{\frac{x+b}{a-b}}\right]^{\frac{x(a-b)}{x+b}}$.Since $\lim _{x \rightarrow \infty} \frac{x(a-b)}{x+b} = a-b$,the limit is $e^{a-b}$.

$\lim _{x \rightarrow \infty} \left(\frac{x+a}{x+b}\right)^{x}$ is equal to

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