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{\bf Question}

Use L'Hopital rule to evaluate the following limits
\begin{description}
\item[(a)]$\ds \lim_{x \to 1} \frac{x^2 - 2x +1}{x^2 - 3x + 2}$
\item[(b)]$\ds \lim_{x \to 0} \frac{\cos x - \cos
2x}{x^2}$
\item[(c)]$\ds \lim_{x \to 1} \frac{\sin \pi x}{\cos \frac{\pi x}{2}}$
\end{description}

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{\bf Answer}

\begin{description}



\item[(a)]
Consider the expression $\displaystyle \frac{x^2 - 2x +1}{x^2 - 3x
+ 2}$. Both numerator and denominator tend to zero as
$x\rightarrow 1.$

Using  L'Hopital's rule gives  $\displaystyle \lim_{x \to 1}
\frac{x^2 - 2x +1}{x^2 - 3x + 2}   =  \lim_{x \to 1} \frac{2x -
2}{2x - 3} = \frac{0}{-1} = 0$




\item[(b)]

Consider the expression $\displaystyle\frac{\cos x - \cos
2x}{x^2}$. Both numerator and denominator tend to zero as
$x\rightarrow 0.$

Using  L'Hopital's rule gives $$\lim_{x \to 0} \frac{\cos x - \cos
2x}{x^2}  =   \lim_{x \to 0} \frac{\frac{d}{dx}\cos x - \cos
2x)}{\frac{d}{dx}(x^2)} \\  = \lim_{x \to 0} \frac{-\sin x + 2
\sin 2x}{2x}$$ Again numerator and denominator both tend to zero
as  $x\rightarrow 0$. We use L'Hopital's rule again to give
$$\lim_{x \to 0} \frac{-\sin x + 2 \sin 2x}{2x}= \lim_{x \to 0}
\frac{-\cos x + 4 \cos 2x}{2}  = \frac{3}{2}$$



\item[(c)]
Consider the expression $\displaystyle\frac{\sin \pi x}{\cos
\frac{\pi x}{2}}$. Both numerator and denominator tend to zero as
$x\rightarrow 1.$

Using L'Hopital's rule gives $$\lim_{x \to 1}\frac{\sin \pi
x}{\cos \frac{\pi x}{2}}  =  \frac{\pi \cos \pi x}{-\frac{1}{2}
\pi \sin \frac{\pi x}{2}} = \frac{\pi \cos \pi }{-\frac{1}{2} \pi
\sin \frac{\pi}{2}} = \frac{-\pi}{-\frac{1}{2} \pi} = 2$$

\end{description}


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