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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?

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    $\begingroup$ It's not true that $\sin^2(x/2)=2\sin(x/2)\cos(x/2)$. $\endgroup$
    – Surb
    Jan 28 at 9:45
  • $\begingroup$ Apologies. The title is wrong. Let me edit it. $\endgroup$
    – Sarthak
    Jan 28 at 9:46
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    $\begingroup$ $x\cot x $ does not tend to $0$ as $x\to 0$. $\endgroup$
    – Koro
    Jan 28 at 9:47
  • $\begingroup$ Oh yeah I got it thank you. But can you please answer my last question? $\endgroup$
    – Sarthak
    Jan 28 at 9:48
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    $\begingroup$ 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. $\endgroup$
    – Koro
    Jan 28 at 10:16
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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.

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  • $\begingroup$ 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. $\endgroup$ Jan 29 at 3:28
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Using your first idea

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

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