# Application with and without L'Hospital's rule differ

I have the limit $$\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h}$$. Using the mean value theorem, we have $$f''(c)$$ for some $$c\in[a-h,a+h]$$. However, we can also use L'Hospital's rule, which reduces the limit to $$\lim_{h\rightarrow0}\frac{f''(a+h)+f''(a-h)}{2}$$, or $$f''(a)$$. I'm not sure why there is a discrepancy, and I'm almost completely certain I didn't make any careless mistakes. We can assume that the first and second derivatives are continuous.

• For $h$ tends to zero, then $c$ tends to $a$ – Dr. Sonnhard Graubner Jun 28 '19 at 17:54

As $$h\rightarrow0 , c\rightarrow a$$ [By squeeze Theorem or Sandwich Theorem ] So there is no discrepancy and they both point to the same thing ie

$$\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h} = f''(a) = f''(c)$$

Also there is a method 3 which can be employed ie $$\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h} =$$

$$\lim_{h\rightarrow0}\frac{(f'(a+h)-f'(a))-(f'(a-h)-f'(a))}{2h} =$$

$$\frac{\lim_{h\rightarrow0}\frac{(f'(a+h)-f'(a))}{h} +\lim_{h\rightarrow0}\frac{(f'(a-h)-f'(a))}{-h}}{2}=$$

$$\frac{f''(a)+ f''(a)}{2}=f''(a)$$

As $$\lim_{h\rightarrow0}\frac{(f(a+h)-f(a))}{h} = f'(a)$$

$$\lim_{h\rightarrow0}\frac{(f(a-h)-f(a))}{-h} = f'(a)$$

The mean value theorem does not say that the limit is $$f''(c)$$, only that there exists $$c\in[a-h,a+h]$$ with the expression equivalent to $$f''(c)$$. The resolution of the problem comes from noticing that $$c$$ and $$f''(c)$$ depend on $$h$$, and then $$\lim_{h\to0}c=a$$ by the squeeze theorem.