# Alternate way to solve $\lim\limits_{x \to 0} (\sin x) ^x$?

My solution:

$$\lim_{x \to 0} (\sin x)^x = \lim_{x \to 0} e^{(x)(\ln \sin x)} = \exp \left( \lim_{x \to 0} (x) (\ln \sin x) \right)$$

Now we have $$\lim_{x \to 0} (x) (\ln\sin x)$$.

Now we can say that the limit is $$0$$ as $$\ln$$, $$\sin x$$ decreases more slowly than $$x$$.

Question: Is there any other method to solve the above limit without using the arguments saying one function decreases slower?

• L’Hôpital’s rule, maybe? Commented Jul 18, 2020 at 11:45
• I guess you meant that $\ln\sin x$ decreases more slowly than $\frac1x$?
– user239203
Commented Jul 18, 2020 at 11:46
• Is there a mathematical definition of ‘decreases more slowly than …’, anyway? Commented Jul 18, 2020 at 19:13

I would write $$x \ln (\sin x)=x \ln (\frac {\sin x} x) +x\ln x$$. The first term tends to $$(0)(0)=0$$ so we only have to find $$\lim x \ln x$$. For this we can apply L'Hopital's Rule to $$\lim \frac {\ln x} {1/x}$$.

• Very intelligent way to solve it 😃. For my humble opinion it is the best answer. Commented Jul 18, 2020 at 11:55
• @Sebastiano Rather obvious since $\sin x\approx x$ for $x\to0$. Commented Jul 18, 2020 at 12:00
• @SameerBaheti Yes, the interesting thing is the approach to write $x \ln (\sin x)=x \ln (\frac {\sin x} x) +x\ln x$. Commented Jul 18, 2020 at 12:01
• @Sebastiano In a competitive exam with a time limit, I would've calculated $\displaystyle\lim_{x\to0}x^x$. I look at the graph of $\sin x$ very close to $0$. But rigor demands such a solution. Commented Jul 18, 2020 at 12:04

You can apply Hôpital's rule 💡:

$$e^{\lim_{x \to 0} \frac{1}{\frac{1}{x}} \ln{\left(\sin{\left(x \right)} \right)}} = e^{\lim_{x \to 0} \frac{\frac{d}{dx}\left(\ln{\left(\sin{\left(x \right)} \right)}\right)}{\frac{d}{dx}\left(\frac{1}{x}\right)}}=e^{\lim_{x \to 0}\left(- \frac{x^{2} \cos{\left(x \right)}}{\sin{\left(x \right)}}\right)}$$

Focus on $$\lim_{x \to 0}\left(- \frac{x^{2} \cos{\left(x \right)}}{\sin{\left(x \right)}}\right)$$ you can apply again the Hôpital's rule having an indeterminate form of type $$(0/0)$$.

• It gives $\lim\limits_{x \to 0} (-x^2) (cot x)$ which again uses the same argument that it one decreases slower. I kinda wanna avoid that. I have edited the question now Anyway this was a valid answer before so +1 Commented Jul 18, 2020 at 11:47
• @HrishabhNayal You have $\cot$ if $\sin(x)\neq 0$. Commented Jul 18, 2020 at 11:52
• Ah! I see so we separate $\frac {x}{sin x}$ to give one and the rest tends to zero Thanks! Commented Jul 18, 2020 at 11:55
• @Sebastiano: isn't $\log \sin x$ a problem for $x \to_{0^{-}}$?
– Alex
Commented Jul 18, 2020 at 12:41
• @Alex My answer to the user is on the limit $\lim_{x \to 0}\left(- \frac{x^{2} \cos{\left(x \right)}}{\sin{\left(x \right)}}\right)$ and not, for the domain, for $\log \sin x$ where $\sin(x)>0$. Commented Jul 18, 2020 at 15:46

A shorter approach using L’Hôpital’s : $$\lim_{x\to 0} x\ln(\sin x) \\ = \lim_{x\to 0} \frac{x}{\sin x} \cdot \sin x \ln(\sin x) \\=\lim_{x\to 0} \sin x \ln(\sin x)$$ Substitute $$\sin x= h$$, $$=\lim_{h\to 0} h\ln h \overset{\text{L.H.}}= 0$$ and so the original limit is $$1$$.

• Very nice also this. 😃 Commented Jul 18, 2020 at 12:00

For $$x\in \left[0,\frac{\pi}2\right]$$ we have

$$\frac 2{\pi}x \leq \sin x \leq x$$

Now, using the standard limit $$\lim_{x\to 0^+}x^x = 1$$ you get

$$1 = \lim_{x\to 0^+}\left(\left(\frac 2{\pi}\right)^xx^x\right)\leq \lim_{x\to 0^+}\sin^x x \leq \lim_{x\to 0^+}x^x = 1$$

• Is $\frac 2{\pi}x \leq \sin x \leq x$ a known identity or is it something obvious I dont see? Also how is $1 = \lim_{x\to 0^+}\left(\left(\frac 2{\pi}\right)^xx^x\right)$ ? Can you explain? Commented Jul 18, 2020 at 12:27
• @HrishabhNayal : These are well known facts. That $\sin x \leq x$ can for example be seen from the Taylor expansion of $\sin x$. That $\frac 2{\pi} x \leq \sin x$ can bee easily seen from the graph and be proved using the concavity of $\sin x$ on the considered interval. The last fact comes from the continuity of $a^x$ at $x=0$. Commented Jul 18, 2020 at 13:13

Just a small variant on Kavi Rama Murthy's answer, together with a comment on the one-sidedness of the limit:

Note,

$$(\sin x)^x=\left(\sin x\over x\right)^xx^x$$

We have

$$\lim_{x\to0}\left(\sin x\over x\right)^x=1^0=1$$

and

$$\lim_{x\to0^+}x^x=1$$

(from the easy L'Hopital for $$x\ln x={\ln x\over1/x}$$). Therefore

$$\lim_{x\to0^+}(\sin x)^x=1$$

Remark: I've specified the limit from the right, $$x\to0^+$$, since $$x^x$$ and $$(\sin x)^x$$ are undefined (as real-valued functions) for (small) negative values of $$x$$. The function $$\left(\sin x\over x\right)^x$$ is defined for negative values of $$x$$ near $$0$$, so it's OK to consider its limit as $$x\to0$$ from both sides.

A short solution using equivalence of functions near $$0$$:

We have $$\sin x\sim_0 x$$, whence $$\: x\ln(\sin x)\sim_0 x\ln x$$, which tends to $$0$$ as $$x \to 0$$. Therefore $$\lim_{x\to0}(\sin x)^x=\lim_{x\to0}\mathrm e^{x\ln(\sin x)}=\mathrm e^0=1.$$