# How do you find the limit of cosx/(1-sinx) as x approaches pi/2+?

Oct 19, 2016

Use L'Hôpital's rule to discover that it approaches infinity as x approaches $\frac{\pi}{2}$

#### Explanation:

If you try to evaluate the limit at $\frac{\pi}{2}$ you obtain the indeterminate form $\frac{0}{0}$; this means that L'Hôpital's rule applies.

To implement the rule, take the derivative of the numerator:

$\frac{d \left\{\cos \left(x\right)\right\}}{\mathrm{dx}} = - \sin \left(x\right)$

take the derivative of the denominator.

$\frac{d \left\{1 - \sin \left(x\right)\right\}}{\mathrm{dx}} = - \cos \left(x\right)$

Assemble this into a fraction:

${\lim}_{x \to \frac{\pi}{2}} \frac{- \sin \left(x\right)}{- \cos \left(x\right)}$

Please observe that the above is the tangent function:

${\lim}_{x \to \frac{\pi}{2}} \tan \left(x\right)$

It is well known that the tangent function approaches infinity as x approaches $\frac{\pi}{2}$, therefore, the original expression does the same thing.

Oct 19, 2016

Multiply by $\frac{1 + \sin x}{1 + \sin x}$

#### Explanation:

$\cos \frac{x}{\left(1 - \sin x\right)} \cdot \frac{\left(1 + \sin x\right)}{\left(1 + \sin x\right)} = \frac{1 + \sin x}{\cos} x$

Now as $x \rightarrow {\left(\frac{\pi}{2}\right)}^{+}$, we have

$1 + \sin x \rightarrow 2$ and

$\cos x \rightarrow {0}^{-}$

So

${\lim}_{x \rightarrow {\left(\frac{\pi}{2}\right)}^{+}} \cos \frac{x}{1 - \sin x} = {\lim}_{x \rightarrow {\left(\frac{\pi}{2}\right)}^{+}} \frac{1 + \sin x}{\cos} x = - \infty$