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}$
Show $\lim_{h \to 0} \frac{f(h)+f(-h)}{h^2}=f''(0)$
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0@Ranabir, may be you'll write an answer to your question? – 2012-08-02
2 Answers
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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0Keep doing you$r$ best – 2012-08-02
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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0That's right. thank you – 2012-08-02