# differences between two sided limits for $\lim_{x\to\frac{\pi}{4}} \frac{1}{\cot{x}-1}$

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 $$\frac10$$

• 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 $-\infty$ – Kaster Feb 24 '13 at 10:10
• yes but cot is positive for pi/4 – phi Feb 24 '13 at 10:52
• you have $\cot \pi/4-1$, don't you? – Kaster Feb 24 '13 at 11:00

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 $$\begin{cases} \cot x - 1 < 0 \quad \text{if} \quad x\to \pi^+/4 \\ \cot x -1 > 0 \quad \text{if} \quad x\to \pi^-/4 \end{cases}$$
$$x \to {\pi\over4}^- \cot x > 1 \implies \cot x - 1 > 0 \implies {1\over{\cot x -1}} > 0$$
$$x \to {\pi\over4}^+ \cot x < 1 \implies \cot x - 1 < 0 \implies {1\over{\cot x -1}} < 0$$