Evaluate $\lim_{x\to\infty} (\sqrt{x+\ln x}-\sqrt{x-\ln x})$ I'm trying to solve the following limit:
$$
\lim_{x\to\infty} (\sqrt{x+\ln x}-\sqrt{x-\ln x} ).
$$
I've tried to simplify it with
$$
\lim_{x\to\infty} \frac{2\ln x}{\sqrt{x+\ln x}+\sqrt{x-\ln x}},
$$
and then I got lost (applying the L'Hoptial rule didn't help much).
 A: Hint:  
$$\sqrt{x+\ln x}+\sqrt{x-\ln x}\ge\sqrt x$$
A: No L'Hospital, just factor out $\sqrt{x}$, and you get the limit, because the rest of denominator tends to 2
A: from your last line $$\frac{2\ln x}{\sqrt{x + \ln x} + \sqrt{x - \ln x}} = \frac{2\ln x}{2\sqrt x} + \cdots \rightarrow 0$$ the reason is $$\lim_{x \to \infty} \frac{\ln x}{\sqrt x} = 0.  $$
A: Change $x=1/t$:
$$
\lim_{t\to0^+}\frac{-2\ln t}{\sqrt{1/t-\ln t}+\sqrt{1/t+\ln t}}=
\lim_{t\to0^+}\frac{-2\sqrt{t}\ln t}{\sqrt{1-t\ln t}+\sqrt{1+t\ln t}}=
$$
Now recall (or prove) that
$$
\lim_{t\to0^+}t^\alpha\ln t=0
$$
for all $\alpha>0$. This is easy writing it as
$$
\lim_{t\to0^+}\frac{\ln t}{t^{-\alpha}}\overset{\mathrm{(H)}}{=}
\lim_{t\to0^+}\frac{1/t}{-\alpha t^{-\alpha-1}}=
\lim_{t\to0^+}\frac{t^\alpha}{-\alpha}=0
$$

Alternative way, using $\sqrt{x}=1/t$
\begin{align}
\lim_{x\to\infty} \sqrt{x+\ln x}-\sqrt{x-\ln x}
&=\lim_{t\to0^+}\frac{\sqrt{1-2t^2\ln t}-\sqrt{1-2t^2\ln t}}{t}\\
&=\lim_{t\to0^+}\frac{(1-t^2\ln t+o(t^2\ln t))-(1+t^2\ln t+o(t^2\ln t))}{t}\\
&=\lim_{t\to0^+}\frac{-2t^2\ln t+o(t^2\ln t)}{t}\\
&=\lim_{t\to0^+}(-2t+o(t\ln t))=0
\end{align}
