The value of operatorname{Lt}_{x rightarrow 0 | vedclass

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

Question

(A) We use the standard limit formula $\operatorname{Lt}_{x \rightarrow a} [f(x)]^{g(x)} = e^{\operatorname{Lt}_{x \rightarrow a} g(x)[f(x)-1]}$ for t

Vedclass · Accepted Answer

$e^2$

Vedclass · Answer

(A) We use the standard limit formula $\operatorname{Lt}_{x \rightarrow a} [f(x)]^{g(x)} = e^{\operatorname{Lt}_{x \rightarrow a} g(x)[f(x)-1]}$ for the indeterminate form $1^{\infty}$.Here, $f(x) = \frac{1+5x^2}{1+3x^2}$ and $g(x) = \frac{1}{x^2}$.As $x \rightarrow 0$, $f(x) \rightarrow 1$ and $g(x) \rightarrow \infty$.Thus, the limit is $e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{1+5x^2}{1+3x^2} - 1 \right)}$.$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{1+5x^2 - (1+3x^2)}{1+3x^2} \right)}$.$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{2x^2}{1+3x^2} \right)}$.$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{2}{1+3x^2}}$.$= e^{\frac{2}{1+0}} = e^2$.

The value of $\operatorname{Lt}_{x \rightarrow 0} \left( \frac{1+5x^2}{1+3x^2} \right)^{\frac{1}{x^2}}$ is

Similar Questions

If $g(x) = \frac{x}{[x]}$ for $x > 2$,then $\lim_{x \rightarrow 2^+} \frac{g(x) - g(2)}{x - 2}$ is equal to

$\lim _{x \rightarrow 0} \frac{\sqrt{11+|x|-6 \sqrt{2+|x|}}}{6-2 \sqrt{2+|x|}} = $

If $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}}} \right)^{2x}} = {e^3},$ then $a$ is equal to

The value of $\lim _{n \rightarrow \infty} \frac{(n !)^{1 / n}}{n}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module