lim(2-x)tan(πx/2)

X->1

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- Oct 20th 2007, 12:54 PMwinterwyrmlets see if you can get this beast
lim(2-x)tan(πx/2)

X->1 - Oct 20th 2007, 01:01 PMCaptainBlack
- Oct 20th 2007, 01:01 PMTD!
Seems rather trivial. Perhaps you mean the following?

$\displaystyle

\mathop {\lim }\limits_{x \to 1} \left( {\left( {1 - x} \right)\tan \left( {\frac{{\pi x}}{2}} \right)} \right)

$ - Oct 20th 2007, 01:05 PMwinterwyrm
sorry, it's actually lim (2-x)^tan(Pi x/2)

as x approaches 1

I tried the formatting from word, and it didn't quite work - Oct 20th 2007, 01:07 PMtopsquark
- Oct 20th 2007, 01:11 PMTD!
That's correct. Does this help you?

$\displaystyle

\mathop {\lim }\limits_{x \to 1} \left( {2 - x} \right)^{\tan \left( {\frac{{\pi x}}{2}} \right)} = \exp \left( {\mathop {\lim }\limits_{x \to 1} \left( {\tan \left( {\frac{{\pi x}}{2}} \right)\log \left( {2 - x} \right)} \right)} \right)

$ - Oct 20th 2007, 01:16 PMtopsquark
- Oct 20th 2007, 01:21 PMwinterwyrm
isn't tan(Pi x/2) indeterminate though? I've tried tons of tricks, and I can't seem to crack any way to get the pi x/2 out of there.

- Oct 20th 2007, 01:24 PMTD!
$\displaystyle

\mathop {\lim }\limits_{x \to 1} \left( {\tan \left( {\frac{{\pi x}}{2}} \right)\log \left( {2 - x} \right)} \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\log \left( {2 - x} \right)}}{{\cot \left( {\frac{{\pi x}}{2}} \right)}}

$

Now you can apply l'Hôpital's rule. - Oct 20th 2007, 01:27 PMtopsquark
- Oct 20th 2007, 01:31 PMwinterwyrm
whoa, you guys are awesome, but do you have to handle different indet forms differently? if so, do you just tackle it by taking the derivative? I derivatived it before using the ln/exp rule, every time, lol. Thanks a million to everyone.

- Oct 20th 2007, 01:33 PMTD!
You can only apply l'Hôpital's rule (so f/g -> f'/g' for the limit) to the indeterminate forms 0/0 or ∞/∞. If you have another indeterminate form (0.∞, 1^∞, ...), you have to reduce it to one of the "allowed" indeterminate forms for l'Hôpital's rule.

- Oct 20th 2007, 01:37 PMCaptainBlack
We are interested in:

$\displaystyle

\lim_{x \to 1} ~(2-x)^ { \tan(\pi x/2)}

$

Put $\displaystyle y=x-1$, then:

$\displaystyle

~(2-x)^{ \tan(\pi x/2)}=(1-y)^{\cot(\pi y/2)}

$

but $\displaystyle \cot(\pi y/2) \approx 2/(\pi y)$ for small $\displaystyle y$

So as $\displaystyle y \to 0, \ \ (1-y)^{\cot(\pi y/2)} \to (1-y)^{\frac{2}{\pi y}}$

and with some jiggery pokery this last should come to $\displaystyle e^{2/pi}$ (it should rearrange into the classical limit form for an exponential)

RonL - Oct 20th 2007, 01:38 PMtopsquark
- Oct 20th 2007, 01:39 PMTD!
Yes, although I prefer the way of rewriting as I did (log(...)/cot(...)) because log is easier to differentiate than 1/log.