# L'hopitals rule, f''(a)

• Kate2010
In summary, the task is to show that limh->0 [(f(a+h)-2f(a)+f(a-h))/h2] = f''(a) where f:R->R is differentiable, a is a real number and f''(a) exists. The student has attempted to use l'hopitals rule and is having difficulty determining the regions of differentiability and continuity for f' in (-2a,2a)/{0} and [-2a,2a]. They are unsure if these regions are allowed when h is tending to 0. They also suggest using the regions [a-2h,a+2h] and (a-2h,a+2h)\{0

## Homework Statement

I have to show that limh->0 [(f(a+h)-2f(a)+f(a-h))/h2] = f''(a) where f:R->R is differentiable, a is a real number and f''(a) exists.

## The Attempt at a Solution

I have applied l'hopitals rule as the question advises and have got to f''(a). However, my problem lies in checking the conditions for l'hopitals rule each time. I'm ok with showing the f(x) (numerator) and g(x) are zero when x=0 for each case. However, I am unsure which region I can take for differentiable and continuous. I thought maybe differentiable on (-2a,2a) \ {0} and continuous on [-2a,2a]. I have used these regions each time. But I am also unsure if I can claim that as f''(a) exists then f' is differentiable in the region (-2a,2a)/{0} and continuous in [-2a,2a]. If f''(a) exists, f' is not necessarily differentiable in this region? Or is it? I wasn't sure what else I could do.

Or would it be better to say continuous in the region [a-2h,a+2h], differentiable in the region (a-2h,a+2h)\{0}. Is this allowed even though h is tending to 0?