Question 370 of 401HardMCQ
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:
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Official solution
\[\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^2} + 2x - 1}}\]
\[ = \mathop {\lim }\limits_{x \to 1} {{({x^2} - 1)si{n^2}(\pi x)} \over {({x^2} - 1){{(x - 1)}^2}}}\]
\[ = \mathop {\lim }\limits_{x \to 1} {{{{\sin }^2}(\pi x)} \over {{{(x - 1)}^2}}}\]
Let x=1+h
\[\therefore\] when x \[\to\] 1 then h \[\to\] 0
\[ = \mathop {\lim }\limits_{h \to 0} {{{{\sin }^2}(\pi (1 + h))} \over {{{(1 + h - 1)}^2}}}\]
\[ = \mathop {\lim }\limits_{h \to 0} {{{{\sin }^2}(\pi h)} \over {{h^2}}}\]
\[ = \mathop {\lim }\limits_{h \to 0} {\pi ^2} \times {{{{\sin }^2}(\pi h)} \over {{{(\pi h)}^2}}}\]
\[ = {\pi ^2} \times 1\]
\[ = {\pi ^2}\] Answer the question to unlock