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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?

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  • $\begingroup$ 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 $-\beta h$. $\endgroup$
    – Paramanand Singh
    May 12 '19 at 9:52
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Marian G.
    May 12 '19 at 7:05
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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 $\alpha f^\prime(x_0)$.

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  • $\begingroup$ If those limits exist. $\endgroup$ May 12 '19 at 22:53
  • $\begingroup$ @martycohen See edit. $\endgroup$
    – J.G.
    May 13 '19 at 5:14

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