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Evaluate the limit for the following if exists. $\;\lim \limits_{x \to 1 } x^{\large\frac{1}{x-1}}$

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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, \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:
$\;\lim \limits_{x \to 1 } x^{\large\frac{1}{x-1}}$ is of the form $1 ^{\infty}$
Step 2:
Let $L=\lim\limits_{x \to 1} x^{\large\frac{1}{x-1}}$
$\log L=\log_c \lim\limits_{x \to 1} x^{\frac{1}{x-1}}=\lim \limits_{x \to 1} \log _e x^{\large\frac{1}{x-1}}$ (by composite function theorem)
Step 3:
$\log_e L= \lim\limits_{x \to 1} \log x^{\frac{1}{x-1}}=\lim \limits_{x \to 1} \log _e x^{\large\frac{1}{x-1}}=\lim \limits{x \to } \large\frac{1}{x-1}$$ \log_c x $
which is of the form $\large\frac {-\infty}{\infty}$
Step 4:
Applying L'Hopital's rule, we have
$\log _e L=\lim \limits_{x \to 1} \large\frac{\Large\frac{1}{x}}{1}$$=1$
Step 5:
Therefore $L=e=>\lim \limits_{x \to 1} x^{\Large\frac{1}{x-1}}$$=e$

 

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

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