The value of lim_{x rightarrow 1} frac{(x^{2} | vedclass

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

Question

(D) Let $L = \lim_{x \rightarrow 1} \frac{(x^{2}-1) \sin^{2}(\pi x)}{x^{4}-2x^{3}+2x-1}$.First,factor the denominator: $x^{4}-2x^{3}+2x-1 = (x^{4}-1)

Vedclass · Accepted Answer

$\pi^{2}$

Vedclass · Answer

(D) Let $L = \lim_{x \rightarrow 1} \frac{(x^{2}-1) \sin^{2}(\pi x)}{x^{4}-2x^{3}+2x-1}$.First,factor the denominator: $x^{4}-2x^{3}+2x-1 = (x^{4}-1) - 2x(x^{2}-1) = (x^{2}-1)(x^{2}+1) - 2x(x^{2}-1) = (x^{2}-1)(x^{2}-2x+1) = (x^{2}-1)(x-1)^{2}$.Substitute this into the limit: $L = \lim_{x \rightarrow 1} \frac{(x^{2}-1) \sin^{2}(\pi x)}{(x^{2}-1)(x-1)^{2}} = \lim_{x \rightarrow 1} \frac{\sin^{2}(\pi x)}{(x-1)^{2}}$.Using the property $\sin(\pi x) = \sin(\pi - \pi x) = \sin(\pi(1-x))$,we get:$L = \lim_{x \rightarrow 1} \left( \frac{\sin(\pi(1-x))}{\pi(1-x)} \cdot \pi \right)^{2} = \pi^{2} \cdot \left( \lim_{x \rightarrow 1} \frac{\sin(\pi(1-x))}{\pi(1-x)} \right)^{2} = \pi^{2} \cdot (1)^{2} = \pi^{2}$.

The value of $\lim_{x \rightarrow 1} \frac{(x^{2}-1) \sin^{2}(\pi x)}{x^{4}-2x^{3}+2x-1}$ is equal to

Similar Questions

$\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}} = $

The limit of the perimeter of the regular $n$-gons inscribed in a circle of radius $R$ as $n \to \infty$ is

$\lim _{x \rightarrow \infty} [x - \log (\cosh x)] = $

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

$\lim _{x \rightarrow \infty}\left(\frac{x+8}{x+1}\right)^{x+5} = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module