# show that $f''(a)=\lim _{h\to 0}\frac{f(a+h)+f(a-h)-2f(a)}{h^2}$ [duplicate]

Let $I\subset \mathbb{R}$ be open interval ,let $f:I\to \mathbb{R}$ be differentiable on $I,$ suppose $f''(a)$ exist at $a \in I$.

show that $$f''(a)=\lim _{h\to 0}\frac{f(a+h)+f(a-h)-2f(a)}{h^2}$$

how can we approach this .i was thinking of MVT.

applying MVT on $[a,a+h]$, we have $$f(a+h)-f(a)=hf'(\alpha_1)$$, where $\alpha_1\in(a,a+h)$

similarly $$f(a)-f(a-h)=hf'(\alpha_2)$$ where $\alpha_2\in(a-h,a)$

subtracting these we get $$\frac{f(a+h)+f(a-h)-2f(a)}{h^2}=\frac{(f'(\alpha_1)-f'(\alpha_2))}{h}$$

now taking limit $h \to0$ both side, can we somehow connect right hand side to $f''(a)$

any hint?? or some other method .i have seen this question already on stackexchange ,but most of them are using taylor expension up to 2nd order, but we are given function is twice differentiable at $x=a$,we donot know about twice differentiability at other points,so how can we use taylor expansion in this case???

## marked as duplicate by Guy Fsone, Matthew Leingang, DonAntonio calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 29 '17 at 19:00

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• Second time in one week or so... – DonAntonio Dec 29 '17 at 19:01
• Try to prove it for polynomials it is enough to prove it with $x^n$. Then any function is approximated with a polynomial – Ameryr Dec 30 '17 at 3:48