L'Hopitals Rule

**worc3247** · April 4th 2011, 05:24 AM

i) Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable, let $a\in\mathbb{R}$ . Suppose that $f$''$(a)$ exists.
Prove that $\displaystyle\lim_{h\to 0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2} = f$''$(a)$ .
ii) Suppose further that $f$''$(a)$ exists for all x and that $f$'''$(0)$ exists. Prove that $\displaystyle\lim_{h\to 0}\frac{4(f(h)-f(-h)-2(f(\frac{h}{2})-f(-\frac{h}{2})))}{h^3}=f$'''$(0)$ .

I realise I need to use L'Hopitals rule at some point but i'm not sure how I can justify the conditions needed to use it. If someone could help me with the first part, I may be able to get the second. Thanks!

**Prove It** · April 4th 2011, 05:35 AM

$\displaystyle \frac{f(a + h) - 2f(a) + f(a-h)}{h^2} \to \frac{0}{0}$ as $\displaystyle h \to 0$ . So you can use L'Hospital's Rule here...

**chisigma** · April 4th 2011, 01:16 PM

Originally Posted by worc3247

i) Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable, let $a\in\mathbb{R}$ . Suppose that $f$''$(a)$ exists.
Prove that $\displaystyle\lim_{h\to 0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2} = f$''$(a)$ .

I realise I need to use L'Hopitals rule at some point but i'm not sure how I can justify the conditions needed to use it. If someone could help me with the first part, I may be able to get the second. Thanks!

The use of l'Hopital rule in the basic definition of second order derivative is not 'fully secure'... better is to derive the second derivative as limit as follows...

$\displaystyle f^{''} (a)= \lim_{h \rightarrow 0} \frac{\frac{f(a+h)-f(a)}{h} - \frac{f(a)-f(a-h)}{h}}{h}= \lim_{h \rightarrow 0} \frac{f(a+h)-2 f(a) + f(a-h)}{h^{2}}$

Kind regards

$\chi$ $\sigma$

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