I'm considering the following limit while looking for asymptotes:
$$\lim_{x\to \infty} \arctan\dfrac{\sqrt{x^2+1}}{x-1}$$
Going the route I initially tried, I divide both parts of the fraction by $x^4$ which would give $\arctan(\infty)$, or $\frac{\pi}{2}$ as the result.
But according to Wolfram, the solution is $\dfrac{\pi}{4}$, which you get by instead taking $x^2$ out of the square root in the numerator, then dividing both numerator and denominator by $x$ which gives $\arctan(1)$, or $\frac{\pi}{4}$.
My question is why my initial approach didn't yield the right result, and why it isn't allowed to do it like I did.
You rather pull $x^2$ from the square root: $$ \sqrt{x^2+1}=\sqrt{x^2\left(1+\frac{1}{x^2}\right)} =x\sqrt{1+\frac{1}{x^2}} $$ (because you can assume $x>0$ as you're evaluating the limit for $x\to\infty$).
Then you similarly pull $x$ from $x-1$ getting $$ x\left(1-\frac{1}{x}\right) $$ then you get $$ \lim_{x\to\infty}\arctan\frac{\sqrt{1+1/x^2}}{1-1/x}=\arctan 1 $$ The method is not that different from yours, but you should beware that $x^4=x^{8/2}=\sqrt{x^8}$.