# find the given limits

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.

• "My attempt : i was thinking about the Reimann sum" - With all due respect, thinking about something without doing anything doesn't usually count as an attempt... Nov 3, 2018 at 10:14
• Is [] the floor function here? Nov 3, 2018 at 10:16
• @YuriyS Greatest Integer Function Nov 3, 2018 at 10:18

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

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

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

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

$$\frac1{(n+1)^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.

• thanks u @Yves Daoust Nov 3, 2018 at 10:27
• @YuriyS: oooops, yes. Of course, this doesn't change the answer.
– user65203
Nov 3, 2018 at 10:32
• Nov 3, 2018 at 10:33
• @Mason: this confirms the formula for this moderate $n$, $1.01\times 1/2=0.505$.
– user65203
Nov 3, 2018 at 10:43