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Given that $f'(x)\ne0$ show that $\lim_{h\to 0}[1/(f(a+h)-f(a))-1/hf'(a)]=-f''(a)/2f'(a)^2$.

By wrriting $f'(a)$ into its limit definitions, LHS seems to be $0$, so how to do this problem?

Thanks.

JSCB
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1 Answers1

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Let
$$ L(h)=\frac{1}{f(a+h)-f(a)}-\frac{1}{hf'(a)}. $$

Then

$$ L(h)=-\frac{f(a+h)-f(a)-hf'(a)}{hf'(a)(f(a+h)-f(a))} $$

By Taylor’s formula we have $\frac{f(a+h)-f(a)-hf'(a)}{h^2} \to \frac{f''(a)}{2}$ when $h\to 0$, so

$$ L(h)=-\frac{f(a+h)-f(a)-hf'(a)}{h^2} \frac{1}{f'(a)}\frac{1}{\frac{f(a+h)-f(a)}{h}} \to \frac{-f''(a)}{2} \frac{1}{f'(a)^2} \ \text{when} \ h\to 0 $$

Ewan Delanoy
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  • Is it possible not to taylor formula? Since this exercise is before the chapter about taylor... – JSCB Nov 17 '13 at 07:46
  • Yes you can avoid Taylor but then you need to use L'Hospital. In order for this to work you need to assume that $f''(a)$ exists and as in question $f'(a) \neq 0$ and nothing more. – Paramanand Singh Nov 17 '13 at 08:04