The function f is differentiable at a. Find $\displaystyle \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}$ I'd assume its something to do with manipulating the definition $\displaystyle \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ but have no idea how.
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Originally Posted by vuze88 The function f is differentiable at a. Find $\displaystyle \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}$ I'd assume its something to do with manipulating the definition $\displaystyle \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ but have no idea how. $\displaystyle \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}=p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}\\$ could you finish it?
so is the answer $\displaystyle 2pf'(a)$
Originally Posted by felper $\displaystyle \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}=p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}\\$ could you finish it? $\displaystyle p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}$ $\displaystyle =p\lim_{h\to 0}\frac{f(a+ph)-f(a)}{ph}+\frac{f(a)-f(a-ph)}{ph}$ $\displaystyle =p\lim_{h\to 0}\frac{f(a+ph)-f(a)}{ph}+\frac{f(a-ph)-f(a)}{-ph}=2pf'(a)$
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