Tricky limit problem involving arctan and ln

Question

I'm having difficulties solving this limit problem:

$$\lim_{x\to\infty}\frac{\arctan\left(\sqrt2(x-\frac{1}{\sqrt2})\right)-\frac{1}{2}\ln\left(2(x-\frac{1}{\sqrt2})^2+1\right)}{2\sqrt2}+\frac{\arctan\left(\sqrt2(x+\frac{1}{\sqrt2})\right)+\frac{1}{2}\ln\left(2(x+\frac{1}{\sqrt2})^2+1\right)}{2\sqrt2}$$

Just looking at it feels daunting. I'm wondering how I could simplify and solve this problem in an easy way.

This is an exercise to help me improve.

Note: the answer given is $$\frac{\pi}{2\sqrt2}$$

Answer 1

Just an expansion of my comment to the OP:

$$\lim_{x\to\infty}\frac{\arctan\left(\sqrt2(x-\frac{1}{\sqrt2})\right)-\frac{1}{2}\ln\left(2(x-\frac{1}{\sqrt2})^2+1\right)}{2\sqrt2}+\frac{\arctan\left(\sqrt2(x+\frac{1}{\sqrt2})\right)+\frac{1}{2}\ln\left(2(x+\frac{1}{\sqrt2})^2+1\right)}{2\sqrt2}=$$

$$\lim_{x\to\infty}\frac{\arctan\left(\sqrt2(x-\frac{1}{\sqrt2})\right)+ \arctan\left(\sqrt2(x+\frac{1}{\sqrt2})\right)}{2\sqrt2}+\lim_{x\to\infty} \frac{\frac{1}{2}\ln\left(2(x+\frac{1}{\sqrt2})^2+1\right) -\frac{1}{2}\ln\left(2(x-\frac{1}{\sqrt2})^2+1\right) }{2\sqrt2}=$$

$$\frac{\frac{\pi}{2}+\frac{\pi}{2}}{2\sqrt2}+\lim_{x\to\infty} \frac{\ln\frac{2(x+\frac{1}{\sqrt2})^2+1}{2(x-\frac{1}{\sqrt2})^2+1} }{4\sqrt2}=$$

$$\frac{\pi}{2\sqrt2}+ \frac{\ln\lim_{x\to\infty} \frac{2(x+\frac{1}{\sqrt2})^2+1}{2(x-\frac{1}{\sqrt2})^2+1} }{4\sqrt2}$$

Now,solve the limit under the logarithm and you're finished.

Answer 2

Let's start solving parts of the exercise to make it simpler.
$arctan(\sqrt2(x - \frac 1{\sqrt 2})) = arctan(\sqrt2\cdot x - 1)$ and $\mathop {\lim }\limits_{n \to \infty} arctan(\sqrt2\cdot x -1 ) = \frac \pi2$ . $arctan(\sqrt2(x + \frac 1{\sqrt 2})) = arctan(\sqrt2\cdot x + 1)$ and $\mathop {\lim }\limits_{n \to \infty} arctan(\sqrt2\cdot x -1 ) = \frac \pi2$ .
Now for the $ln$ part, $\ln{a}-\ln{b} = \ln\frac ab$, and for your limit:
$\frac{1}{2}[\ln(2(x+\frac{1}{\sqrt 2})^2+1) - \ln(2(x-\frac 1{\sqrt 2})^2 + 1)] = \frac12\ln(\frac{2x^2-2\sqrt 2 + 1}{2x^2+2\sqrt2+1})$ after doing the calculations. As $\ln$ converges, $\mathop {\lim }\limits_{n \to \infty}\frac12\ln(\frac{2x^2-2\sqrt 2 + 1}{2x^2+2\sqrt2+1}) = \frac12\ln(\mathop {\lim }\limits_{n \to \infty}\frac{2x^2-2\sqrt 2 + 1}{2x^2+2\sqrt2+1}) = \frac12\ln 1 = 0$.
Adding it all together and dividing by $2\sqrt2$, you'll get $\frac{\pi}{2\sqrt2}$ .