lim _{x rightarrow infty}left[frac{x^2+x+3}{x | vedclass

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

Question

(D) Given,$L = \lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$. This is of the form $1^\infty$. Using the formula $\lim _{x \right

Vedclass · Accepted Answer

$e^2$

Vedclass · Answer

(D) Given,$L = \lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$. This is of the form $1^\infty$. Using the formula $\lim _{x \rightarrow \infty}[1+f(x)]^{g(x)} = e^{\lim _{x \rightarrow \infty} f(x) \cdot g(x)}$,we have: $L = \lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3}{x^2-x+2}-1\right]^x$ $= \lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3-x^2+x-2}{x^2-x+2}\right]^x$ $= \lim _{x \rightarrow \infty}\left[1+\frac{2x+1}{x^2-x+2}\right]^x$ $= e^{\lim _{x \rightarrow \infty} \frac{2x+1}{x^2-x+2} \cdot x}$ $= e^{\lim _{x \rightarrow \infty} \frac{2x^2+x}{x^2-x+2}}$ Dividing numerator and denominator by $x^2$: $= e^{\lim _{x \rightarrow \infty} \frac{2+1/x}{1-1/x+2/x^2}} = e^{\frac{2+0}{1-0+0}} = e^2$.

$\lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$ is equal to

Similar Questions

$\mathop {\lim }\limits_{x \to 1} [x] = $

$\mathop {\lim }\limits_{x \to a} \frac{{{x^2} - {a^2}}}{{x - a}} = $

The function $f(x) = \lim_{n \to \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$ is identical to which of the following functions?

Evaluate the given limit: $\mathop {\lim }\limits_{z \to 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

$\mathop {\lim }\limits_{x \to 0} {(1 - ax)^{\frac{1}{x}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module