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Let $f$ be a function such that $f(0)=0$ and $f$ has derivatives of all order .Show that $$\lim_{h \to 0} \frac{f(h)+f(-h)}{h^2}=f''(0)$$ where $f''(0)$ is the second derivative of $f$ at $0$. I proceed in this way: Note that $$f''(0)=\lim_{h \to 0} \frac{f'(h)-f'(0)}{h}=\lim_{h \to 0} \frac{f'(0)-f'(-h)}{h}$$, by definition of derivative. So, L.H.S $$=\lim_{h \to 0} \frac{f(h)+f(-h)}{h^2}$$[$\frac{0}{0}$ form]$$=\lim_{h \to 0} \frac{f'(h)-f'(-h)}{2h}$$[Applying L'Hospital Rule]$$=\frac{1}{2}[\lim_{h \to 0} \frac{f'(h)-f'(0)}{h}+\lim_{h \to 0} \frac{f'(0)-f'(-h)}{h}]$$$$=\frac{1}{2}[f''(0)+f''(0)]$$$$=f''(0)=R.H.S$$. Can I write ?$$f''(0)=\lim_{h \to 0} \frac{f'(0)-f'(-h)}{h}$$

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why don't you just use Taylor expansion? – Norbert Aug 2 '12 at 15:22
How can I use this Please explain – Argha Aug 2 '12 at 15:23
$$f(h)=f(0)+f'(0)h+0.5f''(0)h^2+o(h^2)$$$$f(-h)=f(0)+f'(0)(-h)+0.5f''(0)(-h)^2+o‌​((-h)^2)$$ – Norbert Aug 2 '12 at 15:25
Is my procedure is wrong? – Argha Aug 2 '12 at 15:27
Consider $$\lim_{h\to0}\frac{\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{h}$$ – J. M. Aug 2 '12 at 15:37
up vote 4 down vote accepted

With Taylor expansion $$L.H.S=\lim_{h \to 0} \frac{f(h)+f(-h)}{h^2}$$$$=\lim_{h \to 0} \frac{f(0)+f'(0)h+0.5f''(0)h^2+o(h^2)+f(0)+f'(0)(-h)+0.5f''(0)(-h)^2+o‌​((-h)^2)}{h^2}$$$$=\lim_{h \to 0} 2\frac{f(0)+0.5f''(0)h^2+o(h^2)}{h^2}$$$$=f''(0)$$

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Thank you for give me support Norbert. I don't know that whenever I can understand my problem I can give answer to my own question – Argha Aug 2 '12 at 16:27
Keep doing your best – Norbert Aug 2 '12 at 16:32

I prefer the Taylor polynomial approach, since it is more informative. But the problem yields readily to L'Hospital's Rule.

Since $f(0)=0$, and our function is continuous, the top approaches $0$. Thus, by L'Hospital's Rule, our limit is equal to $$\lim_{h\to 0}\frac{f'(h)-f'(-h)}{2h}.$$ Since the derivative is continuous, the top approaches $0$, and we can use L'Hospital's Rule again to conclude that our limit is equal to $$\lim_{h\to 0}\frac{f''(h)+f''(-h)}{2}.$$ But the second derivative is continuous, so both $f''(h)$ and $f''(-h)$ have limit $f''(0)$.

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We can apply L'hospital rule if the limit $\frac{0}{0}$ or $\frac{\infty}{\infty}$. here we need to show $f'(h)-f'(-h)=0$ to apply L'Hospital rule – Argha Aug 2 '12 at 17:24
But $f'$ is continuous at $0$, so $\lim_{h\to 0}f'(h)=f'(0)$, and $\lim_{h\to 0}f'(-h)=f'(-0)=f'(0)$. Limits are both $f'(0)$, limit of the difference is therefore $0$. – André Nicolas Aug 2 '12 at 17:29
That's right. thank you – Argha Aug 2 '12 at 17:31

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