Prove that $ \lim_{x \to \infty}f(x) = \infty $ using Squeeze Theorem Let $ f(x) = \int_1^x \frac{\sqrt{t+1}}{t} \text{ d}t$
Prove that $ \lim_{x \to \infty}f(x) = \infty $ using Squeeze Theorem
I can't choose $ g(x) $ and $ h(x) $ to use Squeeze Theorem.
Can someone give me a hint. Thanks
 A: Since for all $x\geqslant1$,
$$2(\sqrt x-1)=\int_1^x\frac{\sqrt t}{t}\,dt\leqslant \int_1^x\frac{\sqrt {t+1}}{t}\,dt = f(x),$$
and $2(\sqrt x - 1)\to \infty$, you get that $f(x)\to\infty$.

You don't really need an $h(x)$ in this case, since you are going to $\infty$. If you insist however you can take $$h(x) =2\sqrt2(\sqrt x-1) =\int_1^x\frac{\sqrt{2t}}{t}\,dt \geqslant f(x),$$
which obviously goes to $\infty$ also.
A: Hint: The integrand is larger than $\dfrac{\sqrt t}{t}.$
A: You don't really need to find two functions, just one function that is smaller than your integrand.
First, notice that,
$$\frac{\sqrt{1+t}}{t}=\sqrt{\frac{1+t}{t^2}}=\sqrt{\frac{1}{t^2}+\frac{1}{t}}$$
Since $t>1$
$$\frac{1}{t^2}+\frac{1}{t}>\frac{2}{t^2}\implies\sqrt{\frac{1}{t^2}+\frac{1}{t}}>\frac{\sqrt{2}}{t}$$
$$$$
Now you might be able to do the rest.
A: How about considering $g(t) = \frac{1}{t}$ is easy to prove that $g(t) \leq \frac{\sqrt{1+t}}{t}$ for all $t \geq 1$. Now consider the function $G(x) = \int_1^x g(t)dt = \log(x)-\log(1)$. By monotonie of the integral it follows that $G(x) \leq f(x)$ And since $\log(x) \to \infty$ as $x \to \infty$ the we are done. I know is not the squeeze Theorem in its pure form but perhaps this helps.
