# How do you use L'hospital's rule to find the limit?

Oct 13, 2014

l'Hopital's Rule

Indeterminate Form 1: $0$/$0$

If ${\lim}_{x \to a} f \left(x\right) = 0$ and ${\lim}_{x \to a} g \left(x\right) = 0$,

then ${\lim}_{x \to a} \frac{f \left(x\right)}{g \left(x\right)} = \setminus {\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$.

ex.) ${\lim}_{x \to 0} \frac{\sin x}{x}$

by differentiating the numerator and the denominator separately,

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

Indeterminate Form 2: $\infty$/$\infty$

If ${\lim}_{x \to a} f \left(x\right) = \pm \infty$ and ${\lim}_{x \to a} g \left(x\right) = \pm \infty$,

then ${\lim}_{x \to a} \frac{f \left(x\right)}{g \left(x\right)} = \setminus {\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$.

ex.) ${\lim}_{x \to \infty} \frac{x}{{e}^{x}}$

by differentiating the numerator and the denominator,

$= {\lim}_{x \to \infty} \frac{1}{{e}^{x}} = \frac{1}{\infty} = 0$

Note: There are other indeterminate forms which can be turned into one of the above forms.

I hope that this was helpful.