Interesting Limit

• Jun 10th 2009, 09:32 AM
Media_Man
Interesting Limit
Evaluate: $\displaystyle \lim_{x\rightarrow 0}\left(\frac{(1 + x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}$
• Jun 10th 2009, 09:49 AM
fardeen_gen
One of the first I had encountered when I started limits.

Spoiler:
$\displaystyle e^{-\frac{1}{2}}$

I leave it to the other members to see if they know how to do it.
HINT:
Spoiler:
Use $\displaystyle a = e^{\ln a}$
• Jun 10th 2009, 10:44 AM
Random Variable
My attempt:

first find $\displaystyle \lim_{x \to 0} \ \ln \Bigg( \Big( \frac {(1+x)^{1 \over x}}{e} \Big)^{1 \over x} \Bigg)$

$\displaystyle = \lim_{x \to 0} \ \frac {1}{x} \Big( \ln (1+x)^{1 \over x} - \ln(e) \Big)$

$\displaystyle = \lim_{x \to 0} \ \frac {1}{x} \Big( \frac {1}{x} \ln (1+x) - 1 \Big)$

$\displaystyle = \lim_{x \to 0} \frac {\ln(1+x)-x}{x^{2}}$

which is the indeterminate form $\displaystyle 0 \over 0$

applying L'Hospital's rule

$\displaystyle = \lim_{x \to 0} \frac {\frac{1}{1+x}-1}{2x}$

applying L'Hospital's rule a second time

$\displaystyle = \lim_{x \to 0} \frac {\frac{-1}{(1+x)^{2}}}{2} = -\frac {1}{2}$

so the limit of the original function is $\displaystyle e^{-\frac{1}{2}}$