Integral $\lim_ {t\to0}\frac{1}{t}\int_t^{2t}\frac{\ln(1+x)}{{ {\sin x}}}\ dx$ How to integrate $$\lim_ {t\to0}\frac{1}{t}\int_t^{2t}\frac{\ln(1+x)}{{ {\sin x}}}\ dx\ ?$$Is it okay to use L'Hospital or this method can't be used if inside the integral its $\frac{0}{0} $. I asked this question before but I wasn't attentive enough and I made some mistakes when writing the limit, my apologies
 A: Since $t\to 0$, you can assume $t<\frac{\pi}4$. Then you get for $x\in[t,2t]$
 the estimate
$$
\frac{\ln(1+t)}{\sin(2t)}\leq \frac{\ln(1+x)}{\sin(x)}\leq \frac{\ln(1+2t)}{\sin(t)}.
$$
This yields
$$
t\frac{\ln(1+t)}{\sin(2t)}\leq \int_t^{2t}\frac{\ln(1+x)}{\sin(x)}\leq t\frac{\ln(1+2t)}{\sin(t)}
$$
and
$$
\lim_{t\to 0}\frac{\ln(1+t)}{\sin(2t)}\leq \lim_{t\to0}\frac1t\int_t^{2t}\frac{\ln(1+x)}{\sin(x)}\leq \lim_{t\to 0}\frac{\ln(1+2t)}{\sin(t)}.
$$
Now you can use L'Hôpital to compute, that both limits are $1$ which gives you the solution.
A: Since
$$\int_t^{2t}\frac{\ln(1+x)}{\sin x}dx=\int_0^{2t}\frac{\ln(1+x)}{\sin x}dx-\int_0^t\frac{\ln(1+x)}{\sin x}dx,$$
we have
\begin{align}
\frac{d}{dt}\left[\int_t^{2t}\frac{\ln(1+x)}{\sin x}dx\right]&=\frac{\ln(1+2t)}{\sin 2t}\cdot 2-\frac{\ln(1+t)}{\sin t}\\
&=\frac{2\ln(1+2t)-2\cos t\ln(1+t)}{\sin 2t}
\end{align}
By LH,
\begin{align}
\lim_{t\to0}\frac{1}{t}\int_t^{2t}\frac{\ln(1+x)}{\sin x}dx&=\lim_{t\to0}\left[\frac{2\ln(1+2t)-2\cos t\ln(1+t)}{\sin 2t}\right]\\
&=\lim_{t\to0}\left[\frac{\frac{4}{1+2t}-\frac{2\cos t}{1+t}+2\sin t\ln(1+t)}{2\cos 2t}\right]\\
&=1
\end{align}
A: Since $$\frac{\ln (1+x)}{\sin x} = 1 + o(1)$$
you have
$$\frac{1}{t} \int_t^{2t}\frac{\ln (1+x)}{\sin x} \ \mathrm{d}x =
\frac{1}{t} \int_t^{2t} (1+o(1)) \ \mathrm{d}x = \frac{1}{t} (t+o(t)) = 1+ o(1)$$
i.e.
$$\lim_{t \to 0} \frac{1}{t} \int_t^{2t}\frac{\ln (1+x)}{\sin x} \ \mathrm{d}x = 1$$
A: Too long for a comment. So let me explain here. Let $f(x) = \frac{\ln(x+1)}{\sin x}$. In the question, we are integrating over a smaller and smaller interval. Since $f$ is bounded in any deleted neighborhood of $0$ with radius smaller than $1$, the numerator goes zero as $t$ goes zero.
I don't need to take $F(t) = \int_0^t f(x) \, dx$. I  could take $F(t) = \int_{a}^t f(x)\, dx$, where $a$ is a negative number greater than -1. Now we can write the numerator as $F(2t) - F(t)$.
By FTC, $F'(t) = f(t)$. Thus,
$$\lim_{t\rightarrow 0} \frac{F(2t)-F(t)}{t} \overset{\text{L'H}}{=} \lim_{t\rightarrow 0} \frac{\frac{d}{dt}[F(2t)-F(t)]}{1}  \overset{\text{ chain}}{=} \lim_{t\rightarrow 0} \frac{2f(2t)-f(t)}{1} =  \lim_{t\rightarrow 0}2f(2t) - \lim_{t\rightarrow 0}f(t) =2-1=1$$
