Test Index

Limits, Continuity and Differentiability Part 2

Section: Mathematics

© examsnet.com

Question : 34 of 100

Marks: +1, -0

The value of

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

[29-Jun-2022-Shift-2]

$\;\frac{\pi^2}{6}$
$\;\frac{\pi^2}{3}$
$\;\frac{\pi^2}{2}$
$\pi^2$

Solution: 👈: Video Solution

S=1+\;\frac{5}{6}+\;\frac{12}{6^2}+\;\frac{22}{6^3}+\dots(1)

\;\frac{1}{6} S=\;\frac{1}{6}+\;\frac{5}{6^2}+\;\frac{12}{6^3}+\dots(2)

S-\;\frac{1}{6} S=1+\;\frac{4}{6}+\;\frac{7}{6^2}+\;\frac{10}{6^3}+\dots

\Rightarrow\;\frac{5S}{6}=1+\;\frac{4}{6}+\;\frac{7}{6^2}+\;\frac{10}{6^3}+\dots\;\text{(3)}\;

Now, multiplying both sides by

\;\frac{1}{6}

, we get

\Rightarrow\;\frac{5S}{36}=\;\frac{1}{6}+\;\frac{4}{6^2}+\;\frac{7}{6^3}+\;\frac{10}{6^4}+\dots

Subtract equation (4) from equation (3), we get

\;\frac{25}{36} S=1+\;\frac{3}{6}+\;\frac{3}{6^2}+\;\frac{3}{6^3}+\dots

\Rightarrow\;\frac{25S}{36}=1+\;\frac{\;\frac{3}{5}}{1-\;\frac{1}{6}}

=1+\;\frac{3}{6} \times \;\frac{6}{5}

=1+\;\frac{3}{5}=\;\frac{8}{5}

\Rightarrow S=\;\frac{8}{5} \times \;\frac{36}{25}=\;\frac{288}{125}

© examsnet.com

Go to Question:

Prev Question

Next Question