# How do you find the limit of  (cos(x)/sin(x) + 1)  as x approaches 0 using l'hospital's rule?

Oct 17, 2017

See below.

#### Explanation:

Plugging in zero gives:

$\frac{1}{0} + 1$

L'Hospital's Rule can only be used when we have a quotient of an indeterminate form:

$\frac{0}{0}$ , $\frac{\infty}{\infty}$

Since we do not have this form, l'hospital's rule cannot be used.

If you evaluate the limit by plugging in values approaching 0 from the left the limit will be:

${\lim}_{x \to {0}^{-}} \cos \frac{x}{\sin} \left(x\right) + 1 = - \infty$

And from the right:

${\lim}_{x \to {0}^{+}} \cos \frac{x}{\sin} \left(x\right) + 1 = \infty$

So the limit is undefined

Using L'Hospital's rule:

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

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

${\lim}_{x \to 0} \left(\frac{- \sin \left(x\right)}{\cos} \left(x\right) + 1\right) = \left(\frac{- \sin \left(0\right)}{\cos} \left(0\right)\right) + 1 = \left(\frac{0}{1} + 1\right) = \textcolor{red}{1}$

color(red)(lim_(x->0^)cos(x)/sin(x) +1 != 1)

So L'Hospital's rule fails.

Oct 17, 2017

${\lim}_{x \to 0} \left(\frac{\cos x}{\sin x} + 1\right)$

$= {\lim}_{x \to 0} \frac{\cos x + \sin x}{\sin x}$

By direct substitution, this gives $\frac{1}{0}$. This is not one of the indeterminate forms of L'Hospital's rule.

${\lim}_{x \to {0}^{-}} \frac{\cos x + \sin x}{\sin x} = - \infty$

${\lim}_{x \to {0}^{+}} \frac{\cos x + \sin x}{\sin x} = \infty$

$\therefore {\lim}_{x \to 0} \frac{\cos x + \sin x}{\sin x} \text{ DNE}$