find the given limits

Question

Suppose that $f$ is twice continuously differentiable on $(-1,1)$ and that $f(0) = 0$.
Find: $$\lim_{x \rightarrow 0^+} \sum_{k=1}^{[\frac{1}{\sqrt x}]}f(kx)$$

My attempt: I was thinking about the Riemann sum, but I am not able to solve.

Any hints/solution will be appreciated.

Thank you.

Answer 1

$$f(x)=f'(0)x+\frac{f''(0)}2x^2+o(x^2)$$

and

$$\lim_{x \rightarrow 0^+} \sum_{k=1}^{[\frac{1}{\sqrt x}]}f(kx)=\lim_{x \rightarrow 0^+} \sum_{k=1}^{[\frac{1}{\sqrt x}]}(f'(0)kx+\frac{f''(0)}2k^2x^2+o(k^2x^2)).$$

If we set $x=\dfrac1{n^2}$,

$$\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\left(f'(0)\frac k{n^2}+\frac{f''(0)}2\frac{k^2}{n^4}+o\left(\frac{k^2}{n^4}\right)\right)$$ the only term that does not vanish is

$$f'(0)\frac{(n+1)n}2\to\frac{f'(0)}2.$$

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For complete rigor, we must account for the fact that for a given $n$,

$$\frac1{(n+1)^2}<x\le\frac1{n^2}$$ and there will be a corrective term bounded by

$$\sum_{k=1}^{n}k\frac{(n+1)^2-n^2}{n^2(n+1)^2},$$ which is harmless.