Applying theorems and properties to solve indeterminate limits I have been trying to solve the following limit for my university Mathematical Analysis class:
$$\lim \limits_{x \to +\infty}\sin(x)(\ln{(\sqrt{x}+1})-\ln(\sqrt{x+1}))$$
I know that the answer is 0, but I am not sure why.
Since $\lim \limits_{x \to +\infty}\sin(x)$ is indeterminate, I immediately discarded substitution as a possibility, especially considering that the logarithms would also be indeterminate with f(x) as a whole being $\sin(\infty)(\ln(\infty)-\ln(\infty))$. I have tried rewriting the problem as $$\lim \limits_{x \to +\infty}\frac{\ln{(\sqrt{x}+1})-\ln(\sqrt{x+1})}{\frac{1}{\sin(x)}}$$ to use L'Hopital theorem, which I simplified to $$\lim \limits_{x \to +\infty}\frac{\frac{1}{2x+2\sqrt{x}}-\frac{1}{2x+2}}{-\frac{\cos(x)}{\sin^2(x)}}$$ But this also seemingly has no solution, and just complicates the problem a lot more that is likely required to solve it. I "know" how to use Landau symbols, so there probably is a way to apply them here, but I have not thought of it yet. I am not aware of any theorems or properties that could help solve this problem. Any help would be appreciated.
 A: The limit calculation can be done in two parts: first the limit without $\sin x$:
$$\begin{align*}
\lim_{x\to+\infty}\left[\ln\left(\sqrt x + 1\right)-\ln\left(\sqrt{x+1}\right)\right]
&= \lim_{x\to+\infty}\ln \frac{\sqrt x+1}{\sqrt{x+1}}\\
&= \lim_{x\to+\infty}\ln \frac{1 + 1/\sqrt x}{\sqrt{1 + 1/x}}\\
&= \ln 1\\
&= 0
\end{align*}$$
Second part, by the squeeze theorem:
$$\begin{array}{rcl}
-1 \le& \sin x &\le 1\\
-\left[\ln\left(\sqrt x + 1\right)-\ln\left(\sqrt{x+1}\right)\right] \le & \sin x \left[\ln\left(\sqrt x + 1\right)-\ln\left(\sqrt{x+1}\right)\right] &\le \left[\ln\left(\sqrt x + 1\right)-\ln\left(\sqrt{x+1}\right)\right]\\
0 \le & \lim_{x\to+\infty}\sin x \left[\ln\left(\sqrt x + 1\right)-\ln\left(\sqrt{x+1}\right)\right] &\le 0
\end{array}$$
So $\lim_{x\to+\infty}\sin x \left[\ln\left(\sqrt x + 1\right)-\ln\left(\sqrt{x+1}\right)\right] = 0$.
A: Asymptotic analysis is very powerful, and you should learn how to use it. Even if it is not permitted on your tests/assignments, it will give you the appropriate insight into how to find and prove the limit value. (Note that we can achieve similar results without asymptotic analysis, but knowing more powerful tools never hurts.)
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
As $x → \infty$:
  $\sqrt{x+1} = \sqrt{x}·\sqrt{1+\lfrac1x} ∈ \sqrt{x}·\big(1+O(\lfrac1x)\big)$.
  Thus $\sin(x)·(\ln(\sqrt{x}+1)−\ln(\sqrt{x+1}))$
    $∈ \sin(x)·\Big(\ln(\sqrt{x}+1)−\ln\big(\sqrt{x}·\big(1+O(\lfrac1x)\big)\big)\Big)$
    $⊆ \sin(x)·\Big(\ln\big(1+\lfrac1{\sqrt{x}}\big)−\ln\big(1+O(\lfrac1x)\big)\Big)$
    $⊆ \sin(x)·\Big(\big(\lfrac1{\sqrt{x}}+O(\lfrac1x)\big)−O(\lfrac1x)\Big)$   [since $\lfrac1{\sqrt{x}} ≈ 0$ and $O(\lfrac1x) ≈ 0$]
    $= \sin(x)·\Big(\lfrac1{\sqrt{x}}+O(\lfrac1x)\Big)$.
Thus not only can you easily find the limit (since $\sin(x) ∈ O(1)$), you know the precise asymptotic behaviour; its dominant term is $\sin(x)·\lfrac1{\sqrt{x}}$.
A: We don't need L'Hopital. Indeed, there is a tag on this site dedicated to evaluating limits with L'Hopital's rule! It can overcomplicate things, and allow you to lose sight of what you're really doing.
With the sometimes simpler "direct" approach, we get a quick solution. $|\sin|\le1$ everywhere, and the natural logarithm is continuous: $$\ln(\sqrt{x}+1)-\ln(\sqrt{x+1})=\ln\left(\frac{\sqrt{x}+1}{\sqrt{x+1}}\right)\to0$$
For all $\varepsilon\gt0$, there exists $M\in\Bbb R$ with $x\gt M\implies|\ln(\sqrt{x}+1)-\ln(\sqrt{x+1})|\lt\varepsilon$. It then follows that: $$\begin{align}|\sin(x)(\ln(\sqrt{x}+1)-\ln(\sqrt{x+1}))|&=|\sin(x)|\cdot|\ln(\sqrt{x}+1)-\ln(\sqrt{x+1})|\\&\le1\cdot|\ln(\sqrt{x}+1)-\ln(\sqrt{x+1})|\\&\lt\varepsilon\end{align}$$For all $x\gt M$. By definition of $\lim_{x\to\infty}$, the above inequality proves the limit is $0$.
Edit: not only do we not need L’Hopital, but on closer inspection of your approach I find that L’Hopital doesn’t even apply anyway! Since $\lim_{x\to\infty}\frac{1}{\sin(x)}$ doesn’t exist, it in particular doesn’t equal $0$, so the $0/0$ rule doesn’t apply.
If you want clarification on the fact that the logarithm term tends to $0$, notice that $\frac{\sqrt{x}+1}{\sqrt{x+1}}\to1$ as $x\to\infty$ and $\ln(1)=0$, hence the continuity of the logarithm gives the result. You can see the ratio of square roots tends to $0$ by L’Hopital (if you wish), by Taylor series (if you wish) or by appealing to the continuity of square root:
$$\begin{align}\lim_{x\to\infty}\frac{\sqrt{x}+1}{\sqrt{x+1}}&=\lim_{x\to\infty}\left(\frac{x+1+2\sqrt{x}}{x+1}\right)^{1/2}\\&=\lim_{x\to\infty}\left(1+2\cdot\frac{1}{x^{1/2}+x^{-1/2}}\right)^{1/2}\\&=(1+0)^{1/2}\\&=1\end{align}$$Where the only tools you need are the facts that $(\cdot)^{1/2}$ is continuous and $\sqrt{x}\to\infty$ as $x\to\infty$: both facts are easy to verify.
