I have been asked to calculate (if exists) the value of $a$ that makes that the following limit exists and is different to $0$.
$$\lim_{x\to0}\frac{\int_0^{x^2}\frac{\log(1+t)-at}{t}dt}{(1-\cos(x/2))^2}$$
My initial idea was to apply L'Hôpital rule and FTC (and then proceed by equivalent infinitesimal), but I can't because the integral is improper of second kind: $\frac{\log(1+t)-at}{t}$ is not continuous at $t=0$.
I ran out of ideas, so any help would be appreciate. Thanks!
Best Answer
Let$$\varphi(t)=\begin{cases}\frac{\log(1+t)-at}t&\text{ if }t\ne0\\1-a&\text{ if }t=0.\end{cases}$$Then $\varphi$ is continuous. Now, let$$F(x)=\int_0^x\varphi(t)\,\mathrm dt.$$Since $\varphi$ is continuous, $F'=\varphi$. So,$$F'(x)=\varphi(x)=1-a-\frac x2+\frac{x^2}3+\cdots$$and so, since $F(0)=0$,$$F(x)=(1-a)x-\frac{x^2}4+\frac{x^3}9+\cdots$$It follows from this that$$\int_0^{x^2}\frac{\log(1+t)-at}t\,\mathrm dt=(1-a)x^2-\frac{x^4}4+\frac{x^6}9+\cdots$$On the other hand,$$\left(1-\cos\left(\frac x2\right)\right)^2=\frac{x^4}{64}-\frac{x^6}{1536}+\cdots$$and so your limit exists if and only if $a=1$.