# How do you find lim sin(2x)/ln(x+1) as x->0 using l'Hospital's Rule?

May 2, 2017

$2$

#### Explanation:

Notice that attempting to evaluate the limit as $x \rightarrow 0$ yields the indeterminate form $\sin \frac{0}{\ln} \left(1\right) = \frac{0}{0}$.

This is in a valid indeterminate form for l'Hopital's rule (the only two valid forms are $\frac{0}{0}$ and $\frac{\infty}{\infty}$).

When one of these indeterminate forms is present, l'Hopital's rule states that:

${\lim}_{x \rightarrow a} f \frac{x}{g \left(x\right)} = {\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

So, we can take the numerator and denominator separately, as per l'Hopital's rule:

${\lim}_{x \rightarrow 0} \sin \frac{2 x}{\ln} \left(x + 1\right) = {\lim}_{x \rightarrow 0} \frac{2 \cos \left(2 x\right)}{\frac{1}{x + 1}} = \frac{2 \cos \left(0\right)}{\frac{1}{0 + 1}} = 2$