# If a = limit (x^2-(ln(x+1))^2)/2x^2, what is a? x go to 0 A.2 B.0 C.5 D.1

### 1 Answer | Add Yours

You need first to evaluate the limit, such that:

`lim_(x->0) (x^2-(ln(x+1))^2)/(2x^2) = (0^2-(ln(0+1))^2)/(2*0^2)`

`lim_(x->0) (x^2-(ln(x+1))^2)/(2x^2) = 0/0`

The indetermination 0/0 requests for you to use l'Hospital's theorem, such that:

`lim_(x->0) (x^2-(ln(x+1))^2)/(2x^2) = lim_(x->0) ((x^2-(ln(x+1))^2)')/((2x^2)')`

`lim_(x->0) ((x^2-(ln(x+1))^2)')/((2x^2)') = lim_(x->0) (2x - 2ln(x+1)*(1/(x+1)))/(4x) = 0/0`

You need to use again l'Hospital's theorem, such that:

`lim_(x->0) (2x - 2ln(x+1)*(1/(x+1)))/(4x) = lim_(x->0) (2 - 2/(x+1)^2 + (2ln(x+1)/(x+1)^2))/4`

`lim_(x->0) (2 - 2/(x+1)^2 + (2ln(x+1)/(x+1)^2))/4 = (2 - 2/(0+1)^2 + (2ln(0+1)/(0+1)^2))/4 = 0/4 = 0`

**Since evaluating the limit yields 0, you need to select the answer B.0.**