The value of lim _{x rightarrow infty}left(fr | vedclass

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

Question

(A) The given limit is of the form $1^{\infty}$.We use the formula $\lim _{x \rightarrow a} [f(x)]^{g(x)} = e^{\lim _{x \rightarrow a} g(x)[f(x)-1]}$.

Vedclass · Accepted Answer

$e^{2}$

Vedclass · Answer

(A) The given limit is of the form $1^{\infty}$.We use the formula $\lim _{x \rightarrow a} [f(x)]^{g(x)} = e^{\lim _{x \rightarrow a} g(x)[f(x)-1]}$.$\lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x} = e^{\lim _{x \rightarrow \infty} x \left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2} - 1\right)}$$= e^{\lim _{x \rightarrow \infty} x \left(\frac{x^{2}-2 x+1 - (x^{2}-4 x+2)}{x^{2}-4 x+2}\right)}$$= e^{\lim _{x \rightarrow \infty} x \left(\frac{2 x-1}{x^{2}-4 x+2}\right)}$$= e^{\lim _{x \rightarrow \infty} \frac{2 x^{2}-x}{x^{2}-4 x+2}}$Dividing numerator and denominator by $x^{2}$,we get:$= e^{\lim _{x \rightarrow \infty} \frac{2 - 1/x}{1 - 4/x + 2/x^{2}}} = e^{2/1} = e^{2}$.

The value of $\lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x}$ is

Similar Questions

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

$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals

$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} = $

$\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module