# Proving $\lim_{h\to 0}\frac{f(x_0+\alpha h)-f(x_0 - \beta h)}{h}=(\alpha + \beta)f'(x_0)$

This question is from Pg 95 of Protter and Morrey's "A First Course in Real Analysis" (1977).

If $$f$$ is differentiable at $$x_0$$, prove that $$\lim_{h\to 0}\frac{f(x_0+\alpha h)-f(x_0 - \beta h)}{h}=(\alpha + \beta)f'(x_0).$$

Since the left-hand side is an indeterminate form of type $$0/0$$, I apply L'Hopital's rule to obtain $$\lim_{h\to 0}\frac{f(x_0+\alpha h)-f(x_0 - \beta h)}{h}=\alpha\lim_{h\to 0}f'(x_0+\alpha h)+\beta\lim_{h\to 0}f'(x_0-\beta h).$$ However, I cannot conclude that the limits on the right-hand side reduce to $$f'(x_0)$$ since the derivative $$f'$$ need not be continuous at $$x_0$$. Is L'Hopital's rule appropriate for this problem?

• L'Hospital's Rule can't be applied here as we don't know if $f$ is differentiable at any point other than $x_0$. Use the fact that $f(x_0+h) =f(x_0)+hf'(x_0)+o(h)$ and replace $h$ by $\alpha h$ and $-\beta h$. May 12, 2019 at 9:52

Use first principles.

$$\lim_{h\to 0}\dfrac{f(x_0+\alpha h)-f(x_0 - \beta h)}{h}=\lim_{h\to 0}\dfrac{f(x_0+\alpha h)-f(x_0)+f(x_0)-f(x_0 - \beta h)}{h}=\lim_{h\to 0}\bigg[\dfrac{f(x_0+\alpha h)-f(x_0)}{\alpha h}\alpha-\dfrac{f(x_0-\beta h)-f(x_0)}{-\beta h}(-\beta)\bigg]=(\alpha + \beta)f'(x_0)$$

As was pointed out in the comments, this is not valid if either of $$\alpha$$ and $$\beta$$ is zero but that should be easy.

• For sake of completeness, one should mention how to handle the simple cases $\alpha =0$ and/or $\beta =0$. For these values your proof strategy cannot be utilized. May 12, 2019 at 7:05

No, for the reason you gave. Instead write your limit as the difference of two, $$\lim_{h\to0}\frac{f(x_0+\alpha h)-f(x_0)}{h}-\lim_{h\to0}\frac{f(x_0-\beta h)-f(x_0)}{h}$$. In particular $$\lim_{h\to0}\frac{f(x_0+\alpha h)-f(x_0)}{h}$$ is $$0$$ if $$\alpha=0$$, or is otherwise $$\alpha\lim_{h\to0}\frac{f(x_0+\alpha h)-f(x_0)}{\alpha h}=\alpha\lim_{k\to0}\frac{f(x_0+k)-f(x_0)}{k}=\alpha f^\prime(x_0)$$, so in either case is $$\alpha f^\prime(x_0)$$.

• If those limits exist. May 12, 2019 at 22:53
• @martycohen See edit.
– J.G.
May 13, 2019 at 5:14