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  4. undersetx arrow 1 textLim ( sin 2(x3+x2+x-3)/1- cos (x2-4 x+3)) has the value equal to

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Q. $\underset{x \rightarrow 1}{\text{Lim}} \frac{\sin ^{2}\left(x^{3}+x^{2}+x-3\right)}{1-\cos \left(x^{2}-4 x+3\right)}$ has the value equal to

Limits and Derivatives

Solution:

$\underset{x \rightarrow 1} {\text{Lim}}\frac{\sin ^{2}\left(x^{3}+x^{2}+x-3\right)}{\left(x^{3}+x^{2}+x-3\right)^{2}} \cdot \frac{\left(x^{3}+x^{2}+x-3\right)^{2}}{1-\cos \left(x^{2}-4 x+3\right)}$
$=(1) \cdot \underset{x \rightarrow 1} {\text{Lim}}\frac{\left(x^{2}-4 x+3\right)^{2}}{1-\cos \left(x^{2}-4 x+3\right)} \cdot \frac{\left(x^{3}+x^{2}+x-3\right)^{2}}{\left(x^{2}-4 x+3\right)^{2}}$
$=(1)(2) \underset{x \rightarrow 1} {\text{Lim}}\left(\frac{x^{3}+x^{2}+x-3}{x^{2}-4 x+3}\right)^{2}=2 l^{2}$
where $l= \underset{x \rightarrow 1} {\text{Lim}} \frac{3 x^{2}+2 x+1}{2 x-4}$ (using L'Hospital's rule)
$=\frac{6}{-2}=-3$
$\therefore L=2(-3)^{2}=18$