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Limits Continuity and Differentiability question

2022 · 29 Jun · Shift 2 · Q26

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Limits Continuity and Differentiability
2022 · 29 Jun · Shift 2 · Q26

Limits Continuity and Differentiability question

2022 · 29 Jun · Shift 2 · Q26

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

The value of

\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^3} + 2x - 1}}

is equal to:

A
${{{\pi ^2}} \over 6}$
B
${{{\pi ^2}} \over 3}$
C
${{{\pi ^2}} \over 2}$
D
$\pi$ 2

View written solutionFree

Correct answer: D

We need to evaluate $L=\lim_{x\to 1}\frac{(x^2-1)\sin^2(\pi x)}{x^4-2x^3+2x-1}.$
First, factor the denominator: $x^4-2x^3+2x-1=(x-1)^3(x+1).$

Also, $x^2-1=(x-1)(x+1).$

So the expression becomes

=\lim_{x\to 1}\frac{\sin^2(\pi x)}{(x-1)^2}.$$ 3. Now rewrite using the standard limit. Let $$t=\pi(x-1).$$ As $x\to 1$, we have $t\to 0$. Also, $$\sin(\pi x)=\sin(\pi+\pi(x-1))=-\sin(\pi(x-1))=-\sin t.$$ Hence, $$\sin^2(\pi x)=\sin^2 t.$$ Therefore, $$L=\lim_{x\to 1}\frac{\sin^2 t}{(x-1)^2}.

Since $t=\pi(x-1)$ , we get $(x-1)^2=\frac{t^2}{\pi^2}.$ So,

=\pi^2\lim_{t\to 0}\left(\frac{\sin t}{t}\right)^2.

Using $\lim_{t\to 0}\frac{\sin t}{t}=1,$ we obtain $L=\pi^2.$

Hence the value of the limit is $\boxed{\pi^2}.$
Comparing with the given options, this corresponds to option D (written as $\pi^2$ ).

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