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For the limit: $$\displaystyle\lim_{x\to\frac{\pi}{4}} \frac{1}{\cot{x}-1}$$ the one sided limits are $\displaystyle\lim_{x\to\frac{\pi}{4}^{-}} \frac{1}{\cot{x}-1}=\infty$

and $\displaystyle\lim_{x\to\frac{\pi}{4}^{+}} \frac{1}{\cot{x}-1}=-\infty$.

Why? should they not be both $+\infty$? I dont really understand how to evaluate limits in which i arrive to $\frac10$

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  • $\begingroup$ If you divide 1 to some extremely small negative number, you'll get extremely big (in absolute value) negative number. If you take limit of it it gets $-\infty$ $\endgroup$ – Kaster Feb 24 '13 at 10:10
  • $\begingroup$ yes but cot is positive for pi/4 $\endgroup$ – phi Feb 24 '13 at 10:52
  • $\begingroup$ you have $\cot \pi/4-1$, don't you? $\endgroup$ – Kaster Feb 24 '13 at 11:00
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You know that $$\lim_{x\to\frac{\pi}{4}} \cot x = 1$$ But if you come to $\pi/4$ from the 'right' (by larger values), $\cot x$ grows from something smaller than 1 and tends to 1 by smaller values. On the contrary, for $x$ coming to $\pi/4$ from the 'left' (smaller values), $\cot x$ decays from bigger values than 1 and tends to 1 by larger values.

Thus you have what you wanted since $$\begin{cases} \cot x - 1 < 0 \quad \text{if} \quad x\to \pi^+/4 \\ \cot x -1 > 0 \quad \text{if} \quad x\to \pi^-/4 \end{cases}$$

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$$x \to {\pi\over4}^- \cot x > 1 \implies \cot x - 1 > 0 \implies {1\over{\cot x -1}} > 0$$

$$x \to {\pi\over4}^+ \cot x < 1 \implies \cot x - 1 < 0 \implies {1\over{\cot x -1}} < 0$$

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