$$\lim_{h\to0}{f(a+ph)-f(a-qh)\over h}$$
I know that the answer is $f'(a)\cdot (p+q)$
but, i have question. Derivative may not continuous. So that answer is right?
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What denotes $$f$$ here? – Dr. Sonnhard Graubner Feb 23 '19 at 13:33
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"Derivative may not consecutive" doesn't mean anything. Please try to explain a bit more. – saulspatz Feb 23 '19 at 14:25
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1I reformatted your post to make it more legible. Please check that I haven't changed the meaning you intend. – saulspatz Feb 23 '19 at 14:27
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@saulspatz I guess "consecutive" should be "continous". Is the answer a multiplication $f'(a)\cdot (p+q)$ ? – Peter Feb 23 '19 at 14:30
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@Peter Yes, I just came to the same conclusion myself. I didn't notice the confusion between $q$ and $g$ though. You must be right. – saulspatz Feb 23 '19 at 14:34
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I'm not good at English. Please understand this point. – 임석빈 Feb 23 '19 at 14:35
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1I changed "consecutive" to "continuous." That must be what you meant to say. – saulspatz Feb 23 '19 at 14:38
2 Answers
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Assuming
- $f$ is differentiable at $a$,
- $p\ne0$,
- $q\ne0$,
your limit can be written $$ \lim_{h\to0}\left(p\frac{f(a+ph)-f(a)}{ph}+q\frac{f(a-qh)-f(a)}{-qh}\right)= pf'(a)+qf'(a)=(p+q)f'(a) $$ There's no need to assume that $f'$ is continuous, it can even just exist at $a$.
Of course the formula extends also for $p=0$ or $q=0$ (or both).
If $g$ is a function for which $\lim_{h\to0}g(h)$ makes sense, then, for $r\ne0$, $$ \lim_{h\to0}g(rh)=\lim_{k\to0}g(k) $$ with easy $\varepsilon$-$\delta$ considerations.
egreg
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Hint : Use $$f(a+ph)-f(a-qh)=(f(a+ph)-f(a))-(f(a-qh)-f(a))$$ to write the given expression as a difference of two expressions and divide the first term by $p$ and the second by $q$ to get $f'(a)$ in both cases.
Peter
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