Limit of an accumulation function(integral involved)

The function $$f(x) = \int_0^x \frac{dt}{t+x}$$ is constant equal to $\ln 2$. If you denote

$$g(x) = \int_{0}^{x}\frac{\cos t^3}{t+x}dt$$ you have (for $x > 0$):

$$\left\vert f(x) - g(x) \right\vert \le \int_0^x \left \vert \frac{1 - \cos t^3}{t} \right\vert dt$$

As the map $h : t \mapsto \frac{1 - \cos t^3}{t}$ can be extended by continuity at zero with $h(t) = 0$, you can conclude that $\lim\limits_{x \to 0} \int_0^x \left \vert \frac{1 - \cos t^3}{t} \right\vert dt = 0$.

Hence the limit you're looking for is equal to $\ln 2$.

Attempt:

MVT for integrals:

1)Let $x>0$;

$I(x):=\displaystyle{\int_{0}^{x}}\frac{\cos^3 t}{t+x}=$

$(\cos^3 s )\displaystyle{\int_{0}^{x}}\frac{1}{x+t}dt$, $s \in [0,x].$

$I(x)=$

$(\cos^3 s) (\log (2x)-\log x)=(\cos ^3 s) \log 2$.

Note $\lim_{x \rightarrow 0^+} s(x)=0$.

Take the $\lim_{ x \rightarrow 0^+} I(x)$;

2) Let $x<0$; Can you finish?

A bit late answer but I think worth mentioning it.

You can also calculate the limit by partial integration and using

• $\lim_{x\to 0^+}x\ln x = 0$

So, $$\int_0^x \underbrace{\frac 1{t+x}}_{u'}\underbrace{\cos t^3}_{v}\;dt=\underbrace{\ln 2x \cdot \cos x^3 - \ln x}_{L(x):=} +\underbrace{3\int_0^x \ln (t+x)t^2\sin t^3 dt}_{I(x):=}$$

For $0<x<\frac 12$ you have $$|I(x)|\leq 3|\ln x |\int_0^x t^2\;dt =|x^3\ln x| \stackrel{x \to 0^+}{\longrightarrow}0$$

For $L(x)$ you have

$$L(x) = \ln 2x \cdot \cos x^3 - \ln x\cdot \cos x^3 + \ln x\cdot \cos x^3 - \ln x $$ $$= \cos x^3\cdot\ln 2 + \ln x \cdot(\cos x^3 - 1)$$ $$= \cos x^3\cdot\ln 2 + x^6\ln x \cdot\frac{\cos x^3 - 1}{x^6}$$ $$\stackrel{x \to 0^+}{\longrightarrow}\ln 2$$