Question

show that for twice differentiable function f,
lim_{h rightarrow 0} (f(a+2h)-2f(a+h)+f(a))/h ^{2} = f'' (a)

Answer 1

its the limit definition of 2nd derivative...
\[f''(a) = \lim_{h \rightarrow 0}\frac{f'(a+h) - f'(a)}{h}\]
where
\[f'(a) = \lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}\]
thus
\[\large f''(a) = \lim_{h \rightarrow 0}\frac{\frac{f(a+2h) - f(a+h)}{h}-\frac{f(a+h)-f(a)}{h}}{h}= \frac{f(a+2h)-2f(a+h)+f(a)}{h}\]

Answer 2

sorry the last was cut off and it should have h^2 on bottom