# What is the discrepancy I am facing in $\lim_{x \to 0} \frac{x \sin x}{1- \cos x}$?

Take a look at the following limit: I am missing something simple but I'm not being able to get it. $$\lim_{x \to 0} \dfrac{x \sin x}{1- \cos x}$$

We can write the denominator as $$2\sin^2x/2$$ and the $$\sin x$$ above it as $$2 \sin x/2\cdot \cos x/2$$ which gives: $$\lim_{x \to 0} \dfrac{x \cos x/2}{\sin x/2}$$ which being in $$0/0$$ form can be operated using L'Hopital's rule. We differentiate and then get $$2$$.

But, if we take another approach and directly use that rule in the first step, we get- $$\lim_{x \to 0} \dfrac{x \cos x + \sin x}{\sin x}$$ Here I tried cancelling the sines which gave- $$\lim_{x \to 0}x \cot x+1=1$$ I know that the first one is right, but I think I am making a very common mistake people make in limits. Is cancelling the sines a wrong step here? In general, what should I do to calculate these limits so that I don't make these mistakes?

• It's not true that $\sin^2(x/2)=2\sin(x/2)\cos(x/2)$.
– Surb
Jan 28 at 9:45
• Apologies. The title is wrong. Let me edit it. Jan 28 at 9:46
• $x\cot x$ does not tend to $0$ as $x\to 0$.
– Koro
Jan 28 at 9:47
• Oh yeah I got it thank you. But can you please answer my last question? Jan 28 at 9:48
• Cancelling sines is not wrong here. Note that $\lim(f(x)+g(x))=\lim f(x) + \lim g(x)$, if both limits on RHS exist.
– Koro
Jan 28 at 10:16

Be careful! In fact, $$\lim_{x \rightarrow 0} x \cot x$$ is also an indeterminant form, so a second application of L'Hopital's rule is necessary: $$\lim_{x \rightarrow 0} \frac{x \cos x + \sin x}{\sin x} = \lim_{x \rightarrow 0} \frac{2 \cos x - x \sin x}{\cos x} = 2$$

I always suggest to try direct substitution. In this case, cot 0 is undefined, so you obtain the indeterminant form $$0 \cdot \cot 0 = 0 \cdot \infty$$ Any time you get an infinity in your answer, you need to be careful.

• Just note that $x\cot x =x/(\tan x)$ which is well known to tend to $1$. Using L'Hospital's Rule unnecessarily should not be encouraged. Jan 29 at 3:28

$$1-\cos(x)=2\sin^2\left(\frac{x}{2}\right).$$ Then $$\frac{x\sin(x)}{2\sin^2(x/2)}=2\cdot \frac{(x/2)^2}{\sin ^2(x/2)}\cdot \frac{\sin(x)}{x}.$$

Using $$\sin(x)\sim x$$ at $$0$$ gives the wished answer.

Using your second idea with l'Hospital

You have that \begin{align*} \lim_{x\to 0}\frac{x\sin(x)}{1-\cos(x)}&=\lim_{x\to 0}\frac{x\cos(x)+\sin(x)}{\sin(x)}\\ &=1+\lim_{x\to 0}\frac{x}{\sin(x)}\cdot \cos(x). \end{align*} Using $$\sin(x)\sim x$$ at $$0$$, allows you to conclude.

Using Taylor

You have that $$\sin(x)=x+o(x)$$ and $$\cos(x)=1-\frac{x^2}{2}+o(x^2)$$

Therefore, $$\frac{x\sin(x)}{1-\cos(x)}=\frac{x^2+o(x^2)}{x^2/2+o(x^2)}=2\cdot \frac{1+o(1)}{1+o(1)}=2+o(1).$$