Question

Solving the limiting value using series expansion.
$\lim_{x\to 0} \dfrac{-ln(x^2+1)+\sin^2(x)}{(cos2x-1)^2}$

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{x\to 0} \rm \: \dfrac{-ln(x^2+1)+\sin^2(x)}{(cos2x-1)^2}$

If we substitute directly x = 0, we get

$\rm \: = \: \dfrac{-ln(1)+\sin^2(0)}{(cos(0)-1)^2}$

$\rm \: = \: \dfrac{ - 0 + 0}{ {(1 - 1)}^{2} }$

$\rm \: = \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle\lim_{x\to 0} \rm \: \dfrac{-ln(x^2+1)+\sin^2(x)}{(cos2x-1)^2}$

Now, we know

$\boxed{ \rm{ \:ln(1 + x) = x - \dfrac{ {x}^{2} }{2} + \dfrac{ {x}^{3} }{3} - \dfrac{ {x}^{4} }{4} + \cdots \: }}$

So,

$\boxed{ \rm{ \:ln(1 + {x}^{2} ) = {x}^{2} - \dfrac{ {x}^{4} }{2} + \dfrac{ {x}^{6} }{3} - \dfrac{ {x}^{8} }{4} + \cdots \: }}$

Again, we know that

$\boxed{ \rm{ \:sinx = x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} + \cdots \: }}$

So,

$\rm \: {sin}^{2}x = {\bigg(x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} + \cdots \:\bigg) }^{2}$

Also,

$\boxed{ \rm{ \:cosx = 1 - \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{4} }{4!} + \cdots \: \: }}$

So,

$\boxed{ \rm{ \:cos2x = 1 - \dfrac{ {4x}^{2} }{2!} + \dfrac{16{x}^{4} }{4!} + \cdots \: \: }}$

Now, on substituting the values, in above expression, we get

$\rm \: = \displaystyle\lim_{x\to 0}\rm \: \dfrac{ - {x}^{2} + \dfrac{ {x}^{4} }{2} - \dfrac{ {x}^{6} }{3} + \dfrac{ {x}^{8} }{4} \cdots + {\bigg(x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} + \cdots \:\bigg) }^{2} }{ {\bigg(1 - \dfrac{4{x}^{2} }{2!} + \dfrac{16 {x}^{4} }{4!} + \cdots \: - 1\bigg) }^{2} }$

$\rm \: = \displaystyle\lim_{x\to 0}\rm \: \dfrac{ - {x}^{2} + \dfrac{ {x}^{4} }{2} - \dfrac{ {x}^{6} }{3} + \dfrac{ {x}^{8} }{4} \cdots + {x}^{2} + \dfrac{ {x}^{6}}{ {(3!)}^{2}} + \cdots - 2\dfrac{ {x}^{4} }{3!} + \cdots }{ {\bigg(- {2x}^{2} + \dfrac{16 {x}^{4} }{4!} + \cdots \: \bigg) }^{2} }$

$\rm \: = \displaystyle\lim_{x\to 0}\rm \: \dfrac{ \dfrac{ {x}^{4} }{2} - \dfrac{ {x}^{6} }{3} + \dfrac{ {x}^{8} }{4} \cdots + \dfrac{ {x}^{6}}{ {(3!)}^{2}} + \cdots - \dfrac{ {x}^{4} }{3} + \cdots }{ {x}^{4} {\bigg(- 2 + \dfrac{16 {x}^{2} }{4!} + \cdots \: \bigg) }^{2} }$

$\rm \: = \displaystyle\lim_{x\to 0}\rm \: \dfrac{ \bigg(\dfrac{{x}^{4} }{2} - \dfrac{ {x}^{4} }{3}\bigg) - \dfrac{ {x}^{6} }{3} + \dfrac{ {x}^{8} }{4} \cdots + \dfrac{ {x}^{6}}{ {(3!)}^{2}} + \cdots \cdots }{ {x}^{4} {\bigg(- 2 + \dfrac{16 {x}^{2} }{4!} + \cdots \: \bigg) }^{2} }$

$\rm \: = \displaystyle\lim_{x\to 0}\rm \: \dfrac{ \dfrac{{x}^{4} }{6}- \dfrac{ {x}^{6} }{3} + \dfrac{ {x}^{8} }{4} \cdots + \dfrac{ {x}^{6}}{ {(3!)}^{2}} + \cdots \cdots }{ {x}^{4} {\bigg(- 2 + \dfrac{16 {x}^{2} }{4!} + \cdots \: \bigg) }^{2} }$

$\rm \: = \displaystyle\lim_{x\to 0}\rm \: \dfrac{ {x}^{4} \bigg(\dfrac{1}{6}- \dfrac{ {x}^{2} }{3} + \dfrac{ {x}^{4} }{4} \cdots + \dfrac{ {x}^{2}}{ {(3!)}^{2}} + \cdots \cdots \bigg)}{ {x}^{4} {\bigg(- 2 + \dfrac{16 {x}^{2} }{4!} + \cdots \: \bigg) }^{2} }$

$\rm \: = \displaystyle\lim_{x\to 0}\rm \: \dfrac{ \dfrac{1}{6}- \dfrac{ {x}^{2} }{3} + \dfrac{ {x}^{4} }{4} \cdots + \dfrac{ {x}^{2}}{ {(3!)}^{2}} + \cdots \cdots}{{\bigg(- 2 + \dfrac{16 {x}^{2} }{4!} + \cdots \: \bigg) }^{2} }$

$\rm \: = \: \dfrac{\dfrac{1}{6} }{ \: \: \: {( - 2)}^{2} \: \: \: \: }$

$\rm \: = \: \dfrac{1}{24}$

Hence,

$\rm\implies \:\boxed{ \rm{ \: \displaystyle\lim_{x\to 0} \rm \: \dfrac{-ln(x^2+1)+\sin^2(x)}{(cos2x-1)^2} = \frac{1}{24} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$