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# Evaluate the limit for the following if exists. $\;\lim\limits_{x \to \infty} \large\frac{\frac{1}{x^{2}}-\large 2\tan^{-1}(\frac{1}{x})}{\Large\frac{1}{x}}$

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$
$\;\lim\limits_{x \to \infty} \Large\frac{\frac{1}{x^{2}}-\Large 2\tan^{-1}(\frac{1}{x})}{\Large\frac{1}{x}}$
Let $\large\frac{1}{x}$$=y As x \to \infty ,y \to 0 Step 2: \;\lim\limits_{x \to \infty} \Large\frac{\frac{1}{x^{2}} -\large 2\tan^{-1}(\Large\frac{1}{x})}{\Large\frac{1}{x}}$$=\lim \limits_{y \to 0} \large\frac{y^2-2 \tan ^{-1}y}{y}$
Which is of the form $\large\frac{0}{0}$
Step 3:
Applying L'Hopital's rule,
$\lim \limits_{x \to \infty}\; \Large\frac{1}{x^2}$$-2\tan^{-1}{\Large\frac{1}{x}}$$=\lim\limits _{y \to 0}\; \large\frac{2y-\Large\frac{2}{1+y^2}}{1}$$=-2$

edited Jul 29, 2013 by meena.p