Question

Solve it
$\lim_{x\to0} \dfrac{-ln(x^2+1)+sin(x^2)}{(cos2x-1)^2}$

Answer 1

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

Given expression is

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

If we substitute directly x = 0, we get

$\rm \: = \: \dfrac{-ln(0+1)+sin(0)}{(cos0-1)^2}$

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

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

which is indeterminant form.

So, Consider again

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

We know,

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

So,

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

Now,

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

So,

$\boxed{ \rm{ \:sin( {x}^{2}) \: = \: {x}^{2} - \dfrac{ {x}^{6} }{3!} + \dfrac{ {x}^{10} }{5!} + \cdots \:}}$

Now,

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

So,

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

So, on substituting all these values, in above expression, we get

$\rm \: = \displaystyle\lim_{x \to 0}\rm \frac{ - {x}^{2} + \dfrac{ {x}^{4} }{2} - \dfrac{ {x}^{6} }{3} + \cdots \: + {x}^{2} - \dfrac{ {x}^{6} }{3!} + \dfrac{ {x}^{10} }{5!} + \cdots \:}{ {\bigg(1 - {2x}^{2} + \dfrac{ {16x}^{4} }{4!} + \cdots \: - 1\bigg) }^{2} }$

$\rm \: = \displaystyle\lim_{x \to 0}\rm \frac{ \dfrac{ {x}^{4} }{2} - \dfrac{ {x}^{6} }{3} + \cdots \: - \dfrac{ {x}^{6} }{3!} + \dfrac{ {x}^{10} }{5!} + \cdots \:}{ {\bigg(- {2x}^{2} + \dfrac{ {16x}^{4} }{4!} + \cdots \: \bigg) }^{2} }$

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

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

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

$\rm \: = \: \dfrac{1}{8}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\displaystyle\lim_{x \to 0}\rm \dfrac{-ln(x^2+1)+sin(x^2)}{(cos2x-1)^2} = \frac{1}{8} \: }}$

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

Additional Information :-

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

Answer 2

Answer:

I hope this helps you and thank you bro.