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Using L'Hopital Rule, evaluate $$ \lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$$

I find this question weired. If we just combine the two terms into one single fraction, we get$$\lim_{x \to 0} {\frac {1-x\cot x} {x^2}}=\frac10=\infty$$

If we follow L'Hopital Rule, this is $\infty-\infty$ form. We follow the following process to convert it into $\frac00$form.$$\infty_1 -\infty_2=\frac 1{\frac 1{\infty_1}}-\frac 1{\frac 1{\infty_2}}=\frac {\frac 1{\infty_2}-\frac 1{\infty_1}}{{\frac 1{\infty_1}}{\frac 1{\infty_2}}}$$

So we will get $$\lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}=\lim_{x \to 0} {\frac {x\tan x-x^2}{x^3\tan x}}$$

If you keep differentiating using the rule you will get rid of the form of $\frac00$ in the third step of differentiation, which give you the answer $1 \over 3$. This method is very tedious. Trust me, you don't want to try.

I am wondering is there a smarter way of solving this question? Thanks.

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    $\begingroup$ Be careful. The first limit is not correct: $x\cot(x)$ does not tend to $0$ as $x$ goes to $0$. $\endgroup$ – AugSB Mar 14 '17 at 9:46
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A single application of L'Hospital is sufficient:

$$\frac {1} {x^2}-\frac {\cot x} {x}=\frac{\sin x-x\cos x}{x^2\sin x}\xrightarrow{\text{L'Hospital}}\frac{x\sin x}{2x\sin x+x^2\cos x}=\frac{\dfrac{\sin x}x}{2\dfrac{\sin x}x+\cos x}\xrightarrow{\text{sinc}}\frac1{2+1}.$$

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    $\begingroup$ To the upvoters: I have completely rewritten, you can change your mind if you don't like the new version :-) $\endgroup$ – Yves Daoust Mar 14 '17 at 10:16
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If you do not want to use L'Hospital, I suppose that Taylor expansions are the good way to go.

Considering $$y=\frac {x\tan( x)-x^2}{x^3\tan (x)}$$ and $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ we then have $$y=\frac{\frac{x^4}{3}+\frac{2 x^6}{15}+O\left(x^7\right)}{x^4+\frac{x^6}{3}+\frac{2 x^8}{15}+O\left(x^9\right)}=\frac{1}{3}+\frac{x^2}{45}+O\left(x^3\right)$$ which shows the limit and also how it is approached.

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  • $\begingroup$ Nice answer. (+1) $\endgroup$ – Olivier Oloa Mar 14 '17 at 10:03
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Note that $$ \lim_{x\to0}x\cot x=\lim_{x\to0}\frac{x}{\sin x}\cos x=1 $$ so $\dfrac{1-x\cot x}{x^2}$ is an indeterminate form $0/0$ at $0$.

You can certainly use l’Hôpital: $$ \lim_{x\to0}\dfrac{1-x\cot x}{x^2} = \lim_{x\to0}\frac{-\cot x+\frac{x}{\sin^2x}}{2x}= \lim_{x\to0}\frac{x-\sin x\cos x}{2x\sin^2x} $$ However, this doesn't seem very inviting, but not hard at all. Observe that you can compute instead $$ \lim_{x\to0}\frac{x-\sin x\cos x}{2x^3}= \lim_{x\to0}\frac{1-\cos^2x+\sin^2x}{6x^2}= \lim_{x\to0}\frac{2\sin^2x}{6x^2} $$ Alternatively you can do $$ \frac{1}{x^2}-\frac{\cot x}{x}= \frac{1}{x^2}-\frac{\cos x}{x\sin x}= \frac{\sin x-x\cos x}{x^2\sin x} $$ which is much nicer: $$ \lim_{x\to0}\frac{\sin x-x\cos x}{x^2\sin x}= \lim_{x\to0}\frac{\sin x-x\cos x}{x^3}\frac{x}{\sin x} $$ Since the limit of the second fraction is $1$, we can just compute $$ \lim_{x\to0}\frac{\sin x-x\cos x}{x^3}= \lim_{x\to0}\frac{x\sin x}{3x^2} $$ which is fairly easy.

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I would keep the fraction as $$\frac{1-x\cot x}{x^2} $$

and use the fact that $(\cot)'(x) = -1-\cot^2x$. In this way the second derivative of $1-x \cot x$ is not too bad to calculate at all.

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A Laurent series is a Taylor expansion with negative exponents. The Laurent expansion for $\cot(x)$ is

$$\cot(x) \approx \frac{1}{x} - \frac{x}{3} - O(x^3) \\ \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{\cot(x)}{x}\right) \rightarrow \frac{1}{x^2} - \frac{1}{x^2} + \frac{1}{3} + O(x^2) = \frac{1}{3}$$

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first :

$$\lim_{x\to 0} \frac{\sin x-x}{x^3}=\frac{-1}{6} \tag{1}$$

$$\lim_{x\to 0} \frac{1-\cos x}{x^2}=\frac{3}{6} \tag{2}$$

$$\lim_{x\to 0} \frac{x}{\sin x}=1 \tag{3} $$

$$\frac{1}{x^2}-\frac{\cot x}{x}=\frac{\sin x-x}{x^3}\times \frac{x}{\sin x}+\frac{1-\cos x}{x^2}\times \frac{x}{\sin x}$$

$$L=\lim_{x\to 0} \frac{1}{x^2}-\frac{\cot x}{x}=\lim_{x\to 0}(\frac{\sin x-x}{x^3}\times \frac{x}{\sin x}+\frac{1-\cos x}{x^2}\times \frac{x}{\sin x})$$ Now using $(1) ,(2),(3)$:

$$L=(\frac{-1}{6}\times 1+\frac{3}{6}\times 1)=\frac{2}{6}$$

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