If f(x) is differentiable at x = a and f'(a) = 0, then find the value of \(\displaystyle\lim_{h\to 0}\frac{f(a+2h^2)-f(a-2h^2)}{h^2}\)
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Write the given expression as a difference of two increments at \(a\): \(\dfrac{f(a+2h^2)-f(a-2h^2)}{h^2}=\dfrac{f(a+2h^2)-f(a)}{h^2}-\dfrac{f(a-2h^2)-f(a)}{h^2}\).
Since \(2h^2\to 0\) as \(h\to 0\), by the definition of the derivative \(\displaystyle\lim_{h\to0}\dfrac{f(a+2h^2)-f(a)}{2h^2}=f'(a)\), so \(\dfrac{f(a+2h^2)-f(a)}{h^2}=2\cdot\dfrac{f(a+2h^2)-f(a)}{2h^2}\to 2f'(a)\).
Similarly, \(\dfrac{f(a-2h^2)-f(a)}{h^2}=-2\cdot\dfrac{f(a-2h^2)-f(a)}{-2h^2}\to -2f'(a)\).
Subtracting, the limit becomes \(2f'(a)-(-2f'(a))=4f'(a)\).
Since \(f'(a)=0\), the value of the limit is \(4\times 0=0\).
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