Well the question is to compute :$$ \lim \limits_{x\to0} (\sin x)^x $$

Set $y = (\sin x)^x$ Taking logarithm of both sides, we have $ \ln y = x \ln (\sin x) $. Applying limits we would get $$ \lim \limits_{x\to0} \ln y =\lim \limits_{x\to0} x \ln (\sin x)$$ $$ \Rightarrow \ln y = \lim \limits_{x\to0} x \times \lim \limits_{x\to0} \ln (\sin x) $$ $$ \Rightarrow \ln y = 0 \times \infty =0$$

$$ \Rightarrow y = e^0 =1$$

Is my deduction correct here?

Or should (must) I manipulate things to $\lim \limits_{x\to0} x \ln (\sin x) = \lim \limits_{x\to0} \frac{ \ln (\sin x)}{\frac{1}{x}}$ and then apply l'Hôpital's rule and then try and put the limits?

Please explain your answer.

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    $\begingroup$ The $0\times \infty$ should have been the tip-off that you're dealing with something indeterminate... $\endgroup$ – J. M. isn't a mathematician Oct 16 '11 at 17:55
  • $\begingroup$ @J. M.:Yes,but why? lets say if you are not allowed to use L'hospitals,then how could we do this without referring to other complex theories? .. also I am much interested to know what could go wrong if I use that .. thanks. $\endgroup$ – Quixotic Oct 16 '11 at 18:02
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    $\begingroup$ Do you know the limit of $\sin(x)\log\sin(x)$ when $x\to0^+$? $\endgroup$ – Did Oct 16 '11 at 18:10
  • $\begingroup$ @Didier Piau:Yes,$O$ isn't?I guess it's same for $x\to0^-$ too ...but I can see your point product of indeterminate....however I don't fully understand it. $\endgroup$ – Quixotic Oct 16 '11 at 18:15
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    $\begingroup$ (1) Once you write $\log\sin(x)$ for $x$ near $0$, you assume that $x>0$ since otherwise $\sin(x)\leqslant0$ and $\log\sin(x)$ does not exist. (2) If you know that the limit of $\sin(x)\log\sin(x)$ is $0$, looking at the limit of $x/\sin(x)$ will get you the result (and there is no indeterminate product here). $\endgroup$ – Did Oct 16 '11 at 18:22

First, a relatively minor point. When $x$ is negative and close to $0$, $\sin x$ is negative, and $(\sin x)^x$ is not defined. Thus, to be precise, we want to calculate $$\lim_{x\to 0^+} (\sin x)^x,$$ the limit as $x$ approaches $0$ from the right.

Second, and more importantly, the assertion $0 \times \infty=0$ is, at the very least, highly suspect. Maybe the following example is too simple, but I hope that it will illustrate the issue. Look at the problem of finding $$\lim_{x\to 0^+} \;(e^{-1/x})^x.$$ This problem is easy, since for $x\ne 0$, $(e^{-1/x})^x=e^{-1}$, so the limit is $e^{-1}$. But let us try to solve the problem in imitation of your proposal.

Let $y=(e^{-1/x})^x$. Then $\ln y= x \ln(e^{-1/x})$. Now we deliberately do not notice that $\ln(e^{-1/x})=-1/x$. Instead, write in imitation of what you did that $\lim_{x\to 0^+}\:x=0$, $\lim_{x\to 0^+} \ln(e^{-1/x})=-\infty$ and "therefore" $\lim_{x\to 0^+} \ln y=0\times(-\infty)=0$. From this we would conclude that $\lim_{x\to 0^+} \;y =e^0=1$, which is false.

A procedure that sometimes gives the wrong answer is not one that we want to use!

Your problem: There are at least three incorrect things about the $0 \times \infty =0$ argument. One of them has been already discussed in detail. Another is that in fact as $x$ approaches $0$ from the right, $\ln(\sin x)$ becomes large negative. The third is that if we decide (with proper justification or not) that $\lim_{x\to 0^+}\; \ln y =0$, the conclusion should be that $\lim_{x\to 0^+} y=e^0=1$.

By way of contrast, the L'Hospital's Rule calculation that you propose works well. We have $$\ln y= \frac{\ln(\sin x)}{\frac{1}{x}}.$$ As $x$ approaches $0$ from the right, the top "approaches" $-\infty$, and the bottom "approaches" $\infty$. We conclude that the limit is equal to $$\lim_{x\to 0^+}\frac{\frac{\cos x}{\sin x}}{\frac{-1}{x^2}},$$ if that limit exists. The expression we want to find the limit of can be rewritten as $-\frac{x^2\cos x}{\sin x}$. The limit is easy to calculate, using the fact that $\frac{x}{\sin x}$ approaches $1$. So we conclude that $$\lim_{x \to 0^+} y=0$$ and therefore $$\lim_{x\to 0^+} (\sin x)^x =1.$$

Note that apart from a little slip at the end, your $0 \times \infty=0$ procedure gave the right answer. However, this was (sort of) accidental. The procedure is theoretically deeply flawed, and one can give many examples, not just the $(e^{-1/x})^x$ example that was discussed in detail, where the procedure gives the wrong answer.

  • $\begingroup$ I love your answers! How precise and to the point and always comes with an example! Thank you very much :) $\endgroup$ – Quixotic Oct 16 '11 at 18:33
  • $\begingroup$ One thing,How could you approached the problem if l'Hôpital's is not allowed? $\endgroup$ – Quixotic Oct 16 '11 at 18:45
  • $\begingroup$ I think @Didier Piau has answered that. Take the $\ln$. Multiplying by $\frac{\sin x}{x}$ does not change the limit. So we want the limit as $u$ approaches $0^+$ of $u\ln u$. Let $u=e^{-z}$. We want $\lim_{z \to \infty} -\frac{z}{e^z}$. There are various tools to show this limit is $0$. $\endgroup$ – André Nicolas Oct 16 '11 at 19:08
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    $\begingroup$ @AndréNicolas, +1 for the knowledge and +100 for willing to share it. $\endgroup$ – NoChance Oct 16 '11 at 20:47

Edited: You can never take limits by parts if one part is 0, and the other goes to infinity.

$$ \lim_{x \rightarrow 0} x \ln(\sin x) = \lim_{x \rightarrow 0} \sin (x) \ln(\sin x) \times \frac{x}{\sin x} $$ This is equal to (by putting $\sin x = t$), $$ \lim_{x \rightarrow 0} \frac{x}{\sin x} \times \lim_{t\rightarrow 0} (t \ln t) $$ assuming that both limits exists and are not of type $0 \times \infty$ or similar.

Now, the first limit is clearly 1, (standard limit), and the second one is 0, after a change of variables, $t \mapsto 1/s.$

Thus, the entire limit is 0.

  • $\begingroup$ $\lim \limits_{x\to0} \frac{\ln(\sin x)}{x} = \lim \limits_{x \to 0} \frac{x \ln(\sin x)}{x \sin x }$ how?! $\endgroup$ – Quixotic Oct 16 '11 at 18:19
  • $\begingroup$ The ratio $(\log\sin(x))/x$ is not involved in the question. $\endgroup$ – Did Oct 16 '11 at 18:25

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