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I'm considering this limit: $$\lim\limits_{x \to 0} \frac{x \cos(x) - \sin(x)}{x \sin^2(x)}$$

If I apply de l'Hôpital's rule I get $-\frac{1}{3}$, while if I immediately simplify the $x$ like this:

$$\frac{\cos(x) - \frac{\sin(x)}{x}}{\sin^2(x)}$$

I get $-\frac{1}{2}$ as result, because I write it as (remembering that $\frac{sin(x)}{x} = 1$):

$$\frac{cos(x) - 1}{x^2}$$

which I simplify it with $-\frac{1}{2}$.

What am I getting wrong?

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    How are you getting $-1/2$? We need to see all of your work to find the error. – 2017-01-02
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    The second rearrangement is an indeterminate form yet – 2017-01-02
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    @TheCount I updated my answer – 2017-01-02
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    You can't replace only a part of an expression with its limit. – 2017-01-02

2 Answers 2

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tl;dr: Your error is in the second step. You substitute $\frac{\sin x}{x}$ by its limit, $1$. But this is getting rid of the low-order terms, which matter.

One way to see it rigorously is to do a polynomial approximation near $0$, that is a Taylor expansion: when $x\to 0$, $$ \frac{\sin x}{x} = 1-\frac{x^2}{6} + o(x^2) $$ so that you neglect a term "behaving like" $\frac{x^2}{6}$. Does it matter? Well, again, near $0$: we have $$ \cos x = 1-\frac{x^2}{2} +o(x^2) \tag{1} $$ so $$\begin{align} \cos x - 1 &= -\frac{x^2}{2} +o(x^2) \tag{2}\\ \cos x - \frac{\sin x}{x} &= 1-\frac{x^2}{2} +o(x^2) - \left(1-\frac{x^2}{6} + o(x^2)\right) = -\frac{x^2}{3} +o(x^2) \tag{3} \end{align}$$ which explains the difference you get between you two methods. The low-order term you (erroneously) dismissed in the second did have a role to play.

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    Thanks indeed. Can you also point me out how to "prettify" the formulas (I suppose it is done with latex)? – 2017-01-02
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    Yes, this website uses MathJax, which is a (watered-down) version of $\LaTeX$ for the Internet. You basically can use all $\LaTeX$ macros (or most) here, see [this page](http://math.stackexchange.com/help/notation) for help. – 2017-01-02
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    @f126ck You can also check the source of all answers to see what the code looks like, for instance by clicking on the "edit" link below. – 2017-01-02
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Hint

Before applying l'Hopital rule, use the equivalence $$\sin(x)\sim x \;(x\to 0).$$ it is much easier.

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    I think that the fundamental problem here (in this question) is to understand why two ways of trying to solve the same problem lead to different results. Are they correct? If yes, how is that possible that the results are different? Answers to questions like these would be more helpful. Of course I'm not saying that this hint couldn't actually be useful for the purposes of the solving the original problem. – 2017-01-02
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    It means you made a mistake in the second. – 2017-01-02
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    @SalahFatima Can you point me out the mistake? – 2017-01-02