Limit of an accumulation function(integral involved) I was asked to evaluate the following limit:
$$
\lim\limits_{x \to 0} \int_{0}^{x}\frac{\cos t^3}{t+x}dt
$$
I've tried to evaluate and indefinite form:
$$
\int\frac{\cos t^3}{t+x}dt
$$
by partial integration assuming that $x=const$. An hour later I got nothing.
 WolframAlpha also can't represent 
this in "standard mathematical functions". 
Now, I'm stuck with it. What am I supposed to try next?
 A: 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 $\ln 2$.
A: 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: 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
$$L(x) = \ln 2x \cdot \cos x^3 - \ln x\cdot \cos x^3 + \ln x\cdot \cos x^3 - \ln x $$
$$= \cos x^3\cdot\ln 2 + \ln x \cdot(\cos x^3 - 1)$$
$$= \cos x^3\cdot\ln 2 + x^6\ln x \cdot\frac{\cos x^3 - 1}{x^6}$$
$$\stackrel{x \to 0^+}{\longrightarrow}\ln 2$$
A: Use the substitution $t=xu$ and the integral changes to $$\int_{0}^{1}\frac{\cos (x^3u^3)}{1+u}\,du$$ The integrand is a continuous function of $x, u$ and hence we can take the limit operation inside the integral. The desired limit is thus $\int_{0}^{1}\frac{du}{1+u}=\log 2$.
On the other hand if you don't want to use this exchange of limit operation with integral then just note that $\cos $ is an even function and hence it is sufficient to consider only $x\to 0^{+}$. Then the integrand lies between $\dfrac{\cos x^3}{1+u}$ and $\dfrac{1}{1+u}$ and hence the integral lies between $(\cos x^3)\log 2$ and $\log 2$. By squeeze theorem the desired limit is $\log 2$.
