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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$e^{a-b}$

Vedclass · Answer

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

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

Similar Questions

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

If $\lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+\ldots+(nk+n)] = 33 \cdot \lim_{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot [1^k + 2^k + 3^k + \ldots + n^k]$,then the integral value of $k$ is equal to $....$

Evaluate the limit: $\mathop {\lim }\limits_{x \to 3} (x+3)$

$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} = $

$\lim _{x \rightarrow 3} \frac{x^3-27}{x^2-9} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module