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

##### 1 Answer

May 2, 2017

#### Explanation:

Notice that attempting to evaluate the limit as

This is in a valid indeterminate form for l'Hopital's rule (the only two valid forms are

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

#lim_(xrarra)f(x)/(g(x))=lim_(xrarra)(f'(x))/(g'(x))#

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

#lim_(xrarr0)sin(2x)/ln(x+1)=lim_(xrarr0)(2cos(2x))/(1/(x+1))=(2cos(0))/(1/(0+1))=2#