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

Evaluate the limit: $\lim_{x \to 1} \left[ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3x^2 + 2x} \right]$

$\lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3} = $

$\mathop {\lim }\limits_{\theta \to \pi /6} \frac{{\cot^2 \theta - 3}}{{\csc \theta - 2}} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {1 - \cos 2x} \right)}^2}}}{{2x\tan x - x\tan 2x}}$ is

$\lim _{x \rightarrow 1} \left( \lim _{y \rightarrow \infty} y \left( (e^x)^{1/y} - 1 \right) \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module