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Evaluate the limit for the following if exists. $\;\lim\limits_{x \to 0+} x^{2}\log_{e}x.$

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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)=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$
  • The conditions for L'Hopital's rule remains unchanged if $\lim\limits_{x \to c} f(x)=\pm \infty $ and $\lim\limits_{x \to c} g(x)=\pm \infty$
  • All indeterminate forms can be reduced t $\large\frac{0}{0}$ or $\large\frac{\infty}{\infty}$
Step 1:
$\;\lim\limits_{x \to 0+} x^{2}\log_{e}x.$ is of the form $0 \times (-\infty)$
Step 2:
$\;\lim\limits_{x \to 0+} x^{2}\log_{e}x.=\lim \limits_{x \to 0+} \Large\frac{\log _c x}{\large\frac{1}{x^2}}$ which is of the form $\large\frac {-\infty}{\infty}$
Step 3:
Applying L'Hopital's rule, we have
$\;\lim\limits_{x \to 0+} x^{2}\log_{e}x.=\lim \limits_{x \to 0+} \large\frac{\Large\frac{1}{x}}{\Large\frac{-2}{x^3}}$
$\qquad=\lim\limits_{x \to 0}\large\frac{-x^2}{2}$$=0$

 

answered Jul 29, 2013 by meena.p
edited Jul 29, 2013 by meena.p
 

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