Find this limit $$\lim_{x\to 0^{+}}\left(\int_{x}^{1}\dfrac{\ln{(1+t)}}{\sqrt{t}}dt+\int_{0}^{x}\dfrac{\sin{2t}}{\sqrt{4+t^2}\int_{0}^{x}(\sqrt{y+1}-1)dy}dt\right)$$
My try: since $$\int_{0}^{x}(\sqrt{y+1}-1)dy=\dfrac{2}{3}(1+x)^{3/2}-x$$
maybe $$\lim_{x\to 0^{+}}\int_{x}^{1}\dfrac{\ln{(1+t)}}{\sqrt{t}}dt$$ exsit?
and $$\lim_{x\to 0^{+}}\int_{0}^{x}\dfrac{\sin{2t}}{\sqrt{4+t^2}\int_{0}^{x}(\sqrt{y+1}-1)dy}dt$$ exsit? then I can't
Note that you can use integration by parts on $\displaystyle\int_x^1 \frac{\ln(1+t)}{\sqrt{t}}\,dt$ with $u=\ln(1+t)$ and $\,dv=\dfrac{1}{\sqrt{t}}\,dt$ to get $$\begin{aligned}\int_x^1\frac{\ln(1+t)}{\sqrt{t}}\,dt &= \left.\left[2\sqrt{t}\ln(1+t)\right]\right|_x^1-2\int_x^1 \frac{\sqrt{t}}{t+1}\,dt\\ &= 2\ln 2 - 2\sqrt{x}\ln(1+x) - 4\int_{\sqrt{x}}^1 \frac{s^2}{s^2+1}\,ds\quad(\text{sub: }s^2=t)\\ &= \ln 4 - 2\sqrt{x}\ln(1+x) - 4 \int_{\sqrt{x}}^1 1-\frac{1}{s^2+1}\,ds\\ &= \ln 4-2\sqrt{x}\ln(1+x)-4\left.\left[s-\arctan s\right]\right|_{\sqrt{x}}^1\\ &= \ln 4 + \pi - 4 -2\sqrt{x}\ln(1+x) +4\sqrt{x}-4\arctan\sqrt{x}\end{aligned}$$ Now we see that $\displaystyle\lim_{x\to 0^+}\int_x^1\frac{\ln(1+t)}{\sqrt{t}}\,dt = \ln 4 + \pi -4$.
The other part of the limit can be rewritten as $$\lim_{x\to 0^+}\dfrac{\displaystyle\int_0^x\dfrac{\sin 2t}{\sqrt{4+t^2}}\,dt}{\displaystyle\int_0^x\sqrt{y+1}-1\,dy}\rightarrow \frac{0}{0}$$ Now apply L'Hopital's rule to get that $$\begin{aligned}\lim_{x\to 0^+}\dfrac{\displaystyle\int_0^x\dfrac{\sin 2t}{\sqrt{4+t^2}}\,dt}{\displaystyle\int_0^x\sqrt{y+1}-1\,dy} &= \lim_{x\to 0^+}\dfrac{\dfrac{\sin(2x)}{\sqrt{4+x^2}}}{\sqrt{x+1}-1}\\ &= \lim_{x\to 0^+}\frac{\sin 2x}{\sqrt{x+1}-1}\cdot\lim_{x\to 0^+}\frac{1}{\sqrt{4+x^2}}\\ &= \ldots\end{aligned}$$ Can you take things from here?
In the end, I get $\displaystyle\lim_{x\to 0^+}\frac{\sin 2x}{\sqrt{x+1}-1}=4$; thus, $\displaystyle\lim_{x\to 0^+}\dfrac{\displaystyle\int_0^x\dfrac{\sin 2t}{\sqrt{4+t^2}}\,dt}{\displaystyle\int_0^x\sqrt{y+1}-1\,dy}=4\cdot\frac{1}{\sqrt{4}} = 2$ and therefore $\displaystyle \lim_{x\to 0^+}\left(\int_x^1 \frac{\ln(1+t)}{\sqrt{t}}\,dt+\displaystyle\int_0^x\dfrac{\sin 2t}{\sqrt{4+t^2}\int_0^x\sqrt{y+1}-1\,dy}\,dt\right)=\ln4+\pi-2$
