Evaluating $\lim\limits_{x\to\infty}\int_0^1\frac{\ln x}{\sqrt{x+t}}dt$ Question: Evaluate the limit of this integral

Attempt:

I got the integral. but for some reason I can't find the limit at all. L'Hopital's doesn't even work
 A: Hint:$$\lim_{x \to \infty} \ln x \int_{0}^{1}\dfrac{1}{\sqrt{t+x}}\mathrm dt=2\cdot\lim_{x \to \infty}\dfrac{\sqrt{1+x}-\sqrt{x}}{1/\ln x}=2\cdot \lim_{x\to \infty}\dfrac{\ln x}{\frac{1}{\sqrt{1+x}-\sqrt{x}}}$$

$$\dfrac{1}{\sqrt{1+x}-\sqrt{x}}\cdot\dfrac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}=\sqrt{x+1}+\sqrt{x} $$

$$2\cdot\lim_{x\to \infty} \dfrac{\ln x}{\sqrt{x+1}+\sqrt{x}}=2\lim_{x\to \infty}\dfrac{1/x}{1/2\cdot\left(\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{x}}
\right)}$$
A: $$ \int\dfrac{\log( x)}{\sqrt{t+x}}\, dt=2 \sqrt{t+x} \log (x)$$
$$ \int_{0}^{1}\dfrac{\log( x)}{\sqrt{t+x}}\, dt=2 \left(\sqrt{x+1}-\sqrt{x}\right) \log (x)=2\sqrt{x}\left(\sqrt{1+\frac{1}{x}}-1 \right)\log(x)$$
Now, since $x$ is large
$$\sqrt{1+\frac{1}{x}}=1+\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$
$$2\sqrt{x}\left(\sqrt{1+\frac{1}{x}}-1 \right)\log(x)=2\sqrt{x}\left(\frac{1}{2 x}+O\left(\frac{1}{x^2}\right) \right)\log(x)\sim\frac{\log(x)}{\sqrt x}= 2\frac{\log(\sqrt x)}{\sqrt x}=2\frac {\log(y)}y$$
Try with $x=100$; the exact result would be $2 \left(\sqrt{101}-10\right) \log (100)\approx 0.459371$ while the approximation gives $\frac{\log (10)}{5}\approx 0.460517$.
A: Clearly if $0<t<1$ and $x>1$ then $$\frac{1}{\sqrt{x+1}}<\frac{1}{\sqrt{x+t}}<\frac{1}{\sqrt{x}}$$ Integrating the above with respect to $t$ on interval $[0,1]$ and multiplying the resulting inequality by $\log x$ we get for $x>1$ $$\frac{\log x} {\sqrt{x+1}}<\int_{0}^{1}\frac{\log x} {\sqrt{x+t}} \, dt<\frac{\log x} {\sqrt{x}} $$ Now applying squeeze theorem we get the desired limit as $0$.
