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

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Limits

The value of

lim_{x \to 1} \frac{( x ^{2} - 1 ) s i n ^{2} ( π x )}{x ^{4} - 2 x ^{3} + 2 x - 1}

is equal to:

Answer: (D)

Factor the denominator:

x^{4} - 2 x^{3} + 2 x - 1 = (x^{2} - 1) (x - 1)^{2}

. The limit becomes

lim_{x \to 1} \frac{( x ^{2} - 1 ) s i n ^{2} ( π x )}{( x ^{2} - 1 ) ( x - 1 ) ^{2}} = lim_{x \to 1} \frac{s i n ^{2} ( π x )}{( x - 1 ) ^{2}}

. Let

u = x - 1

, so

x = 1 + u

and as

x \to 1

u \to 0

. Then

sin (π x) = sin (π (1 + u)) = sin (π + π u) = - sin (π u)

. Thus

\frac{s i n ^{2} ( π x )}{( x - 1 ) ^{2}} = \frac{s i n ^{2} ( π u )}{u ^{2}} = π^{2} (\frac{s i n ( π u )}{π u})^{2} \to π^{2}

u \to 0

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option

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