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# Evaluate the limit for the following if exists. $\;\lim\limits_{x \to 2}\large\frac{x^{n}-2^{n}}{x-2}$

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 $\lim\limits_{x \to c}\; f(x)=0, \lim \limits_{x \to c}\; g(x)=0$ and
• $\lim\limits_{x \to c} \large\frac{f'(x)}{g'(x)}$$=L it follows that • \lim \limits_{x \to c} \large\frac{f(x)}{g(x)}$$=L$
Step 1:
$\;\lim\limits_{x \to 2} \large\frac{x^2-2^n}{x- 2}$ is of the form $\large\frac{0}{0}$
Step 2:
Applying L'Hopital's rule,
$\lim \limits_{x \to 2}\; \Large\frac{\tan x^n- 2^n}{x-2}$$=\lim\limits _{x \to 2}\; \Large\frac{nx^{n-1}}{1}$
$\qquad=n.2^{n-1}$

edited Jul 29, 2013 by meena.p