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

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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:
$\;\lim\limits_{x \to 0 +} x^{x}$ is of the form $0^0$
Step 2:
Let $L=\lim\limits _{x \to 0+} x^x$
=> Let $L=\log _e \lim \limits_{x \to 0+} x^x$
$\qquad =\lim \limits_{x \to 0+}\log _e x^x$ (By composite function theroem)
Step 3:
$\log _e L => \lim \limits_{x \to 0+} x \log_ e x$
which is of the form $0 \times (- \infty)$
Now $\log _e L => \lim \limits_{x \to 0+} \large\frac{\log _e x}{\Large\frac{1}{x}}$
which is of the form $\large\frac {-\infty}{\infty}$
Step 4:
Applying L'Hopital's rule, we have
$\log _e L = \lim \limits_{x \to 0+} \large\frac{\Large\frac{1}{x}}{\Large\frac{-1}{x^2}}$
$\qquad = \lim \limits_{x \to 0+} -x=0$
Step 5:
$\therefore L=1\;or\; \lim \limits_{x \to 0+} x^x=1$

 

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

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