Solve this

To Evaluate:

$\lim _{x \rightarrow 1}\left(\frac{x^{4}-3 x^{2}+2}{x^{3}-5 x^{2}+3 x+1}\right)$

L'Hospital's rule

Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval I except at a point a where

$\lim _{x \rightarrow a} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})=0$ or $\pm \infty$

then

$\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\lim _{x \rightarrow a} \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}$

As $x \rightarrow 0$, we have

$\lim _{x \rightarrow 1}\left(\frac{x^{4}-3 x^{2}+2}{x^{3}-5 x^{2}+3 x+1}\right)=\frac{0}{0}$

Therefore,

$\lim _{x \rightarrow 1}\left(\frac{x^{4}-3 x^{2}+2}{x^{3}-5 x^{2}+3 x+1}\right)=\lim _{x \rightarrow 1} \frac{4 x^{3}-6 x}{3 x^{2}-10 x+3}=\frac{4-6}{3-10+3}=-\frac{2}{-4}=-\frac{1}{2}$

Hence

$\lim _{x \rightarrow 1}\left(\frac{x^{4}-3 x^{2}+2}{x^{3}-5 x^{2}+3 x+1}\right)=-\frac{1}{2}$