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Evaluate the limit for the following if exists,$\;\lim\limits_{x \to 2} \large\frac{\sin\pi x}{2-x}$

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Toolbox:
  • L'Hopital's rule: Let $f$ and $g$ be continous real valued functions defined on the closed interval $[a,b], f,g$ be differentiable on $(a,b)$ and $g'(c) \neq 0$
  • Then if $ \lim\limits_{x \to c} \;f(x)=v_1 \lim \limits_{x \to c} g(x)=0$ and
  • $ \lim\limits_{x \to c} \;\large\frac{f'(x)}{g'(x)}$$=L_1$ it follows that
  • $ \lim\limits_{x \to c} \large\frac{f(x)}{g(x)}$$=L$
Step 1:
$\lim \limits_{x \to 2} \large\frac{\sin \pi x}{2-x}$ is of the type $\large\frac{0}{0}$
Step 2:
Applying L'Hopital's rule,
$\lim \limits_{x \to 2} \large\frac{\sin \pi x}{2-x}$$=\lim _{x \to 2} \large\frac{\pi \cos \pi x}{-1}$$=-\pi$

 

answered Jul 26, 2013 by meena.p
 

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