Question

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

Answer 1

`2`

Explanation:

Notice that attempting to evaluate the limit as `xrarr0` yields the indeterminate form `sin(0)/ln(1)=0/0`.

This is in a valid indeterminate form for l'Hopital's rule (the only two valid forms are `0/0` and `oo/oo`).

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`