Question

Evaluate the limit:
$\underset{x\to 1}{lim}{\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^{\frac{1-\cos (x-1)}{(x-1{)}^2}}$

Answer 1

Given:
$\underset{x\to 1}{lim}{\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^{\frac{1-\cos (x-1)}{(x-1{)}^2}}$

Then we have,

$\underset{x\to 1}{lim}{\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^L\cdots \left(1\right)$

Let $L=\frac{1-\cos (x-1)}{(x-1{)}^2}$
As $x\to 1$
It becomes $\left(\frac{0}{0}\right)$

$\underset{x\to 1}{lim}L=\underset{x\to 1}{lim}\frac{2{\sin }^2\left(\frac{x-1}{2}\right)}{4\times \left(\frac{(x-1{)}^2}{4}\right)}$
$[\because 1-\cos 2\theta =2{\sin }^2\theta ]$

$=\frac{1}{2}{lim}_{x\to 1}{\left(\frac{\sin \left(\frac{x-1}{2}\right)}{\left(\frac{x-1}{2}\right)}\right)}^2$

From equation $\left(1\right)$, we get

$\underset{x\to 1}{lim}{\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^{\frac{1}{2}\underset{x\to 1}{lim}{\left(\frac{\sin \left(\frac{x-1}{2}\right)}{\left(\frac{x-1}{2}\right)}\right)}^2}$

$={\left(\frac{1^3+2(1{)}^2+1+1}{1^2+2\left(1\right)+3}\right)}^{\frac{1}{2}\times 1^2}$

$[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$={\left(\frac{5}{6}\right)}^{\frac{1}{2}\times 1^2}=\sqrt{\frac{5}{6}}$

Therefore,
$\underset{x\to 1}{lim}{\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^{\frac{1-\cos (x-1)}{(x-1{)}^2}}=\sqrt{\frac{5}{6}}$