Limit of a function to the power infinite

**cosmonavt** · November 30th 2011, 08:09 AM

What exactly is

Limit of a function to the power infinite-lim.png

?
The possible answers are

(a) $e$
(b) $e^2$
(c) $e^3$
(d) $\frac{1}{e}$

**Amer** · November 30th 2011, 08:45 AM

$\lim _{x \rightarrow 0} e^{\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)^\frac{1}{x}$

since e is continuous
$e^{\lim _{x \rightarrow 0}\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)^\frac{1}{x}$

$\lim _{x \rightarrow 0}\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)^\frac{1}{x}$

$\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)$

$\lim _{x \rightarrow 0}\left(\frac{\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)}{x}\right)$
0/0
use lopital rule

**cosmonavt** · December 3rd 2011, 12:08 AM

Thanks. Some problems though...

How exactly do you make e the base in the second step? (after "since e is continuous")?

**Siron** · December 3rd 2011, 01:52 AM

Originally Posted by cosmonavt

Thanks. Some problems though...

How exactly do you make e the base in the second step? (after "since e is continuous")?

In general, $u=e^{\ln(u)}$ , because $y=\ln(u) \Leftrightarrow e^{y}=u \Leftrightarrow e^{\ln(u)}=u$ . So for this limit, let $u=\tan\left(\frac{\pi}{4}+x\right)^{\frac{1}{x}}$

**cosmonavt** · December 3rd 2011, 03:45 AM

Originally Posted by Siron

In general, $u=e^{\ln(u)}$ , because $y=\ln(u) \Leftrightarrow e^{y}=u \Leftrightarrow e^{\ln(u)}=u$ . So for this limit, let $u=\tan\left(\frac{\pi}{4}+x\right)^{\frac{1}{x}}$

Actually that was not my question. I know that $u=e^{\ln(u)}$ but how did you make e the base in this:

$e^{\lim _{x \rightarrow 0}\ln \left( \tan \left( \frac{\pi}{4}+x \right)\right)^\frac{1}{x}$

**Amer** · December 3rd 2011, 07:40 AM

since the exponential function is continuous
lim f(g(x)) = f(lim g(x))
if f is continuous at lim g(x)

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