Question

Let \(f^{\prime \prime}(x)\) be continuous. Show that $$ \lim _{h \rightarrow 0} \frac{f(x+h)-2 f(x)+f(x-h)}{h^{2}}=f^{\prime \prime}(x) $$

Short answer

From the Taylor series expressions of \(f(x+h)\) and \(f(x-h)\), and through simplification and cancellation, the equation yields: \(\lim _{h \rightarrow 0} \frac{f(x+h)-2 f(x)+f(x-h)}{h^{2}} = f''(x)\), which is the required proof.

Step-by-step solution

Write the Taylor Series

Let's first express \(f(x+h)\) and \(f(x-h)\) using Taylor series up to the second derivative. The Taylor series of \(f(x+h)\) at point x yields:\[f(x+h) = f(x) + hf'(x) + \frac{h^2}{2}f''(x) + o(h^2).\]Similarly, the Taylor series of \(f(x-h)\) at point x is:\[f(x-h) = f(x) - hf'(x) + \frac{h^2}{2}f''(x) + o(h^2). \]

Substitution

Next, substitute the expressions of \(f(x+h)\) and \(f(x-h)\) into the given equation: \[\lim _{h \rightarrow 0} \frac{f(x+h)-2 f(x)+f(x-h)}{h^{2}} = \lim _{h \rightarrow 0} \frac{(f(x) + hf'(x) + \frac{h^2}{2}f''(x) + o(h^2)) - 2f(x) + (f(x) - hf'(x) + \frac{h^2}{2}f''(x) + o(h^2))}{h^{2}}.\]

Simplification & Cancellation

Simplify the numerator of the fraction by cancelling terms:This gives, \[\lim _{h \rightarrow 0} \frac{h^2 f''(x) + o(h^2)}{h^{2}}.\]

Apply Limit

Finally, apply the limit to the equation:As \[h \rightarrow 0, o(h^2) \rightarrow 0.\]Therefore, \[\lim _{h \rightarrow 0} \frac{h^2 f''(x) + o(h^2)}{h^{2}} = f''(x).\]