Limit with L'Hospital with infinite indeterminate formats

I'm trying to find the limit: $$\large \lim_{x\to0}(\sin x)^x$$ Whst I did was apply L'Hospital Rule: $$\large \text{let }y =(\sin x)^x\implies \ln y=x\ln\sin x$$ $$\large \lim_{x\to0}\ln y = \lim_{x\to0} x\ln\sin x = \lim_{x\to0}\frac x{\frac1{\ln\sin x}} = \lim_{x\to0} \frac1{\frac {-\cos x}{(\ln\sin x)^2\sin x}} = \lim_{x\to0} (-\tan x )(\ln\sin x)^2 = \lim_{x\to0} \frac{(\ln\sin x)^2}{\frac1{(-\tan x )}} = ...$$ Ultimately it keeps on going, please help me.

HINT:

$$\lim_{x\to0^+}\frac{\ln\sin x}{\dfrac1x}=\lim_{x\to0^+}\frac{\dfrac{\cos x}{\sin x}}{-\dfrac1{x^2}}=-\lim_{x\to0}x\cdot \lim_{x\to0}\cos x\cdot\frac1{\lim_{x\to0}\dfrac{\sin x}x}$$

Hope you can take it home from here?

if you have $\frac{\log \sin x}{\frac{1}{x}}$ and apply L'Hospital's rule, the log will disappear, so you got to fix it

Whgen evaluating the limit of a product with l'Hopital's rule, I'd advise to put in the denominator the simplest factor. So: \begin{align} \lim_{x\to0} x\ln\sin x&=\lim_{x\to0}\frac{\ln\sin x}{1/x}\\ &=-\lim_{x\to0}x^2\cot x\\ &=\lim_{x\to0}\frac{x^2}{\tan x}\\ &=\lim_{x\to0}\frac{2x}{1+\tan^2x}\\ &=0 \end{align}

• but wouldn't putting x in numerator cancel it for once and always – RE60K Jun 20 '14 at 17:53

Hint: $$\sin x\sim_0 \, x \ \ \Rightarrow \ \ \lim_{x\to0}\:(\sin x)^x=\lim_{x\to0}\:x^x.$$

• the problem is that you get $x^x$, which is also indeterminate and requires logging to get a sensible solution, i.e. L'Hospital's type thing. So why do the whole first thing when you can log apply L'H straight away? – Alex Jun 20 '14 at 17:41
• @Alex I thought that the limit $\lim\limits_{x\to0}x^x=1$ is famous enough. – Hakim Jun 20 '14 at 17:42
• I'm sure it rivals Pythagoras in fame) but it still is an indeterminate form and requires extra steps to get solved. – Alex Jun 20 '14 at 17:43
• @Alex Yes, but since it is famous we don't have to justify its value. IMO it's up to the OP to clarify whether he can use it or not, only then your point may be justified. – Hakim Jun 20 '14 at 17:45

Let the desired limit be $L$ then we have \begin{aligned}\log L &= \log\left(\lim_{x \to 0^{+}}(\sin x)^{x}\right)\\ &= \lim_{x \to 0^{+}}\log(\sin x)^{x}\text{ (by continuity of log)}\\ &= \lim_{x \to 0^{+}}x\log(\sin x)\\ &= \lim_{x \to 0^{+}}x\log\left(\frac{\sin x}{x}\cdot x\right)\\ &= \lim_{x \to 0^{+}}x\log\left(\frac{\sin x}{x}\right) + x\log x\\ &= 0\cdot \log 1 + \lim_{x \to 0^{+}}x\log x\\ &= -\lim_{y \to \infty}\frac{\log y}{y}\text{ (putting }y = 1/x)\\ &= 0\end{aligned} so that the desired limit $L = e^{0} = 1$. Here we have used two standard limits $$\lim_{x \to 0}\frac{\sin x}{x} = 1,\,\,\lim_{x \to \infty}\frac{\log x}{x^{a}} = 0$$ for any $a > 0$. We really don't need powerful tools like L'Hospital for such common problems.

• A better way to solve! – RE60K Jun 22 '14 at 5:14
• @Aditya: I suggest that you also have a look at my blog posts paramanands.blogspot.com/2013/11/… where I try to present step by step approach to common limit problems. – Paramanand Singh Jun 22 '14 at 5:17
• I just read all what you taught about limit(possibly except the proofs) at your blog. I am greatly impressed by your way, and now I think all that my teacher taught me was so different; I need to do it all over again; do you suggest any book for theory/questions? – RE60K Jun 22 '14 at 6:19
• @Aditya: I don't know about any recent books as I have been out of touch. But I mentioned Hardy's "A Course of Pure Mathematics" in my blog. This is the best book for sure, but it needs good effort on part of student. You should check out the review of the book at paramanand.blogspot.com/2005/11/… – Paramanand Singh Jun 22 '14 at 6:23
• I think other answers have given the way to use L'Hospital and you learn that too. I only try not to use LHR for simple problems because LHR is at a conceptually higher level (involves differentiation and can't be used by a beginner who is learning limits but does not know about differentiation). – Paramanand Singh Jun 22 '14 at 6:25