# Thread: lets see if you can get this beast

1. ## lets see if you can get this beast

lim(2-x)tan(πx/2)
X->1

2. Originally Posted by winterwyrm
lim(2-x)tan(πx/2)
X->1
$\lim_{x \to 1_+} (2-x) \tan(\pi x/2)=-\infty$

$\lim_{x \to 1_-} (2-x) \tan(\pi x/2)=+\infty$

RonL

3. Seems rather trivial. Perhaps you mean the following?

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

4. 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

5. Originally Posted by winterwyrm
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
My calculator gives the answer as $e^{2/\pi}$, but I can't tell you why.

-Dan

$
\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)
$

7. Originally Posted by TD!

$
\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)
$
I always forget about that trick.

-Dan

8. 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.

9. $
\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.

10. Originally Posted by winterwyrm
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.
Yes, but it's an indeterminate of the form $\infty \cdot 0$. Do you know how to handle that?

-Dan

11. 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.

12. 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.

13. Originally Posted by winterwyrm
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
We are interested in:

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

Put $y=x-1$, then:

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

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

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

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

RonL

14. Originally Posted by TD!
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.
Such as $\lim_{x \to 1} \frac{tan(\pi x/2)}{\frac{1}{ln(2 - x)}}$...

-Dan

15. Yes, although I prefer the way of rewriting as I did (log(...)/cot(...)) because log is easier to differentiate than 1/log.

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