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$$\lim_{x\to\infty} \cot(1/x) - x$$

I've been rewriting this to $\displaystyle{\lim_{x\to\infty}\frac{\cos\frac{1}{x} - x \sin\frac{1}{x}}{\sin\frac{1}{x}}}$ and then using L'Hopital's, but I just can't seem to get to a result.

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Let $\frac {1}{x}=u $ hence limit changes to $\lim_{u \to 0} \frac {u\cos(u)-\sin(u)}{\sin(u)} $ which is a $\frac {0}{0} $ form. Now use Lhospital thrice to get the answer as $0$.

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Hint

introducing $t=\frac1x$ gives you$$\lim_{x\to\infty}\cot\left(\frac1x\right)-x = \lim_{t\to 0^+} \cot(t)-\frac 1t$$

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With $u=\dfrac1x$ then $$\lim_{u\to0}\dfrac{u\cos u-\sin u}{u\sin u}=\lim_{u\to0}\dfrac{-\sin u}{\frac{\sin u}{u}+\cos u}=0$$ by l'Hopital.

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