TITLE: Calculate $\lim_{x\to 0} \frac{1}{x^2}\int^{1-\cos x}_{\sqrt[3]{ x^7}} \ln (2+\tan^2 t)\,dt$
QUESTION [5 upvotes]: Calculate $$\lim _{x\rightarrow 0}\frac{1}{x^2} \int\limits^{1-\cos x}_{\sqrt[3]{ x^7}} \ln (2+\tan^2 t)\,dt$$

My solution:
Let: $$f(x)=\int^{1-\cos x} _{\sqrt[3]{ x^7}} \ln (2+\tan^2 t)\,dt$$ $$g(x)=x^2$$For $x\rightarrow 0$ we have 
$f,g \rightarrow 0$. So from L'Hôpital's rule: 
$$\lim_{x\rightarrow 0} \frac{f(x)}{g(x)}=\lim_{x\rightarrow 0} \frac{f'(x)}{g'(x)}=\lim_{x\rightarrow 0} \frac{\ln (2+ \tan ^{2} (1-\cos x)) \cdot \sin x -\ln (2+\tan^{2} (x^{\frac{7}{3}})) \cdot \frac{7}{3} \cdot x^{\frac{4}{3}})}{2x}=\ln 2$$
And then I have a question: how do I prove that I can use L'Hôpital's rule? I think that I should say that for $x\rightarrow 0$ for $f$ I have an integral from $0$ to $0$ so $f(x)=0$. But I need a professional explanation so I am asking for a detailed justification of why I can use this rule.

REPLY [2 votes]: Let $u=x^{7/3},v=1-\cos x$ so that both $u, v$ tend to $0$ with $x$. Let the integrand be denoted by $f(t) $ so that $f(t) =\log(2+\tan^2t)$ and $f(0)=\log 2$.
It is unnecessary to use L'Hospital's Rule and one can use the Fundamental Theorem of Calculus more efficiently here. The expression under limit can be written as $$\frac{1}{x^2}\int_{u}^{v}f(t)\,dt=\frac{v}{x^2}\cdot\frac{1}{v}\int_{0}^{v} f(t) \, dt-\frac{u} {x^2}\cdot\frac{1}{u}\int_{0}^{u}f(t)\,dt$$ Next note that $u/x^2\to 0,v/x^2\to 1/2$ and therefore by FTC we have the desired limit as $$\frac{1}{2}\cdot f(0)-0\cdot f(0)=\frac{1}{2}\log 2$$

The usage of L'Hospital's Rule is fine here because both numerator (integral) and denominator ($x^2$) tend to $0$ and these functions are differentiable near $0$, but its use is somewhat complicated and unnecessary here.