# How do you find the limit of  (acota)/(sina)  as a approaches 0?

##### 2 Answers
Nov 9, 2016

Rewriting with $\cot a = \cos \frac{a}{\sin} a$:

$L = {\lim}_{a \rightarrow 0} \frac{a \cot a}{\sin} a = {\lim}_{a \rightarrow 0} \frac{a \cos a}{\sin} ^ 2 a$

This is in the indeterminate form $\frac{0}{0}$, so we can apply L'Hopital's rule and take the derivatives of the numerator and denominator individually:

$L = {\lim}_{a \rightarrow 0} \frac{\frac{d}{\mathrm{da}} \left(a \cos a\right)}{\frac{d}{\mathrm{da}} {\sin}^{2} a} = {\lim}_{a \rightarrow 0} \frac{\cos a - a \sin a}{2 \sin a \cos a}$

This is now no longer an indeterminate form, since it is the form $\frac{1}{0}$. Thus, the limit does not exist (there's an asymptote at $a = 0$).

Nov 9, 2016

Non-existing, $\pm \infty$, disclosing antipodal-infinite discontinuity.

#### Explanation:

$\frac{a \cot a}{\sin} a$=(a/sin a)(cos a/sin a)=(cos a)(csc a)

The are two approaches $\to 0 , {0}_{+} \mathmr{and} {0}_{_}$

$\lim a \to {0}_{+} \csc a = \infty \mathmr{and} \lim a \to {0}_{_} \csc a = - \infty$

But, either way, both (a /sina) and cos a to 1.

So, the limit of the product = product of the limits

$= \left(1\right) \left(1\right) \left(\pm \infty\right)$ and can be symbolized as non-exiting

$\pm \infty$ that reveals antipodal-infinite discontinuity at a = 0..