Question

Solve the limiting value using series expansion.
$\lim\limits_{x\to 0}\dfrac{-\ln(x^2+1) + \sin(x^2)}{(\cos(2x) - 1)^2}$

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{- ln(x^{2} + 1) + sin(x^{2} )}{[cos(2x) - 1]^{2} }$

As we know that,

Put the value of x = 0 in the equation and check their indeterminant form, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{-ln((0)^{2} + 1) + sin((0)^{2} )}{[cos(2(0)) - 1]^{2} } = \dfrac{0}{0}$

As we can see that,

It is in the form of 0/0 indeterminant.

As we know that,

Expansion of :

$\sf \implies \displaystyle ln(1 + x) = x - \frac{x^{2} }{2} + \frac{x^{3} }{3} - . . . . .$

$\sf \implies \displaystyle sin(x) = x - \frac{x^{3} }{3!} + \frac{x^{5} }{5!} - . . . . .$

$\sf \implies \displaystyle cos(x) = 1 - \frac{x^{2} }{2!} + \frac{x^{4} }{4!} - . . . . .$

Using this expansions in the equation, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \frac{- \bigg(x^{2} - \dfrac{x^{4} }{2} + \dfrac{x^{6}}{3} - . . . . . \bigg) + \bigg(x^{2} - \dfrac{x^{6} }{3!} + . . . . . \bigg) }{\bigg(1 - \dfrac{(2x^{2} )^{2} }{2!} + \dfrac{(2x^{2} )^{4} }{4!} - . . . . . - 1\bigg)^{2} }$

$\sf \implies \displaystyle \lim_{x \to 0} \frac{- x^{2} + \dfrac{x^{4} }{2} + x^{2} }{ \bigg(\dfrac{(4x^{4} )^{} }{2!} \bigg)^{2} }$

$\sf \implies \displaystyle \lim_{x \to 0} \bigg[ \dfrac{(x^{4}/2) }{[(4x^{4} /2!)^{2} } \bigg]$

$\sf \implies \displaystyle \lim_{x \to 0} \bigg[\dfrac{1}{4 \times 2} \bigg] = \dfrac{1}{8}$

$\sf \implies \displaystyle \boxed{\lim_{x \to 0} \dfrac{- ln(x^{2} + 1) + sin(x^{2} )}{[cos(2x) - 1]^{2} } = \frac{1}{8} }$