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I'm preparing for a calculus exam and came across this limit:

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

Of course $\tan(\pi/2)$ is undefined, but the excercise was a multiple choice with options $\infty$, 0, e, $e^\pi$ and $e^{2/\pi}$

So WolframAlpha says $\tan(\pi/2)$ is $\infty$, I figured this is a $1^\infty$ indeterminate and first tried to rewrite it like this:

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

And I have $$\lim_{x \to 1}{(1+(1-x))}^{(1-x)^{-1}}=e$$

But then I got stuck trying to calculate $$\lim_{x \to 1}(1-x)\tan(\pi/2)x$$

I've also tried other methods but everything I try I end up stuck in a similar situation. Also, all the ways I can think of to calculate this limit seem too far-fetched (including the one above).

So, is the excercise wrong? If it's undefined, is it because of $\tan(\pi/2)$ or is there some other argument?

David K
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GnP
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4 Answers4

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When it says $\tan(\pi/2)x$, it's meant to be interpreted as $\tan((\pi/2)x)$, not $(\tan(\pi/2))x$ (yes, it is terrible notation). So the expression $(x-1)\tan((\pi/2)x)$ is a $0\cdot\infty$ indeterminate form as $x\to 1$ and you can try using L'Hopital's rule. If it were $(\tan(\pi/2))x$, the expression would of course be undefined for all values of $x$, and so it would be meaningless to talk about the limit.

(If this confusion is not what you are actually asking about, apologies and I will delete this answer!)

Eric Wofsey
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  • Oh, that explains it. I had never seen that before. Then again, I'm pretty new to all this. Thanks! – GnP Oct 08 '15 at 21:23
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    @gnp Probably you had never seen this before because few people would write $\tan(\pi/2)x$ when they really mean $\tan((\pi/2)x)$. It would be fair to say that the problem was written incorrectly, but we make our best guess as to what the problem writer thought it should mean and we answer accordingly. – David K Oct 08 '15 at 21:30
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You may write

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

So lets concentrate on the following limit

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

You may use the Hopital rule to get

$$\mathop {\lim }\limits_{x \to 1} \frac{{\ln \left( {2 - x} \right)}}{{\cot \left( {\frac{\pi }{2}x} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{ - 1}}{{2 - x}}}}{{ - \left( {1 + {{\cot }^2}\left( {\frac{\pi }{2}x} \right)} \right)\frac{\pi }{2}}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {2 - x} \right)\left( {1 + {{\cot }^2}\left( {\frac{\pi }{2}x} \right)} \right)\frac{\pi }{2}}} = \frac{2}{\pi }$$

So the final answer is

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

and we are done.

Hosein Rahnama
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If we take the log of this function: $$lim_{x\rightarrow 1} ln(2-x)^{tan\frac{\pi}{2}x} = lim_{x\rightarrow 1} tan(\frac{\pi}{2}x)ln(2-x)$$ $$= lim_{x\rightarrow 1} \frac{ln(2-x)}{cot(\frac{\pi}{2}x)}$$ By L'Hopitals rule: $$= lim_{x\rightarrow 1} \frac{\frac{-1}{2-x}}{{-csc^2(\frac{\pi}{2}x)*\frac{\pi}{2}}}$$ $$= \frac{\frac{-1}{2-1}}{{-csc^2(\frac{\pi}{2})*\frac{\pi}{2}}}$$ $$= \frac{2}{\pi}$$ But his was the log of our function, so if we raise it to the power e the answer is: $$lim_{x\rightarrow 1} (2-x)^{tan\frac{\pi}{2}x} = e^{\frac{2}{\pi}}$$

Mark
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Try taking logs, interchanging the limit with the log (this can be done since $\log x$ is a continuous function for $x>0$) and then it may become more obvious.

In other words, let $L=\lim_{x\to1}(2-x)^{\tan\left(\frac{\pi}{2}x\right)}$ and then take logs of both sides.

To get you started, I have

$\log L = \log\lim_{x\to1}\left[(2-x)^{\tan\left(\frac{\pi}{2}x\right)}\right] = \lim_{x\to1}\log\left[(2-x)^{\tan\left(\frac{\pi}{2}x\right)}\right]$.

See if you can use limit and log laws to get something plausible.