Evaluating $\lim_{x\to 0}\frac1{x^2}({\arctan(1+x^2)-\arcsin\frac{\cos x}{\sqrt{2}}})$

Question

I've been struggling for a while with the following limit: $$L = \lim_{x\to 0}\frac{\arctan(1+x^2) - \arcsin\left(\frac{\cos x}{\sqrt{2}}\right)}{x^2}$$

I know that $L = 1$, since it seems obvious from the graph and WolframAlpha confirms it, yet I simply can't find a way to theoretically evaluate it. I've tried all sorts of trig substitutions, two of which seem pretty intuitive:

• $\cos x = \sin\alpha+\cos\alpha = \sqrt{2}\sin\left(\alpha + \frac{\pi}{4}\right)$ (suitable for the right part)
• $1+x^2 = \tan\left(\theta + \frac{\pi}{4}\right)$ (suitable for the left part)

I thought of those because the limit kind of hints at them and both new variables also tend to $0$ and I feel I'm really close, but I couldn't finish anything and also started doubtingthis might not be the way to go. Any ideas are really welcome and greatly appreciated!

Edit: I forgot the mention, this limit is somehow supposed te be evaluable without using L'Hopital's rule, that's what all the fuzz is about! :D I also grew suspicious if it's even possible, but it's an interesting challenge nonetheless!

Answer 1

Since $\arctan(1+x^2)-\tfrac{\pi}{4}=\arctan\tfrac{x^2}{2+x^2}\sim\tfrac{x^2}{2}$ and$$\tfrac{\pi}{4}-\arcsin\tfrac{\cos x}{\sqrt{2}}=\arcsin\tfrac{\sqrt{2-\cos^2x}-\cos x}{2}\sim\tfrac{\sin^2x}{\sqrt{2-\cos^2x}+\cos x}\sim\tfrac{x^2}{2},$$the limit is $\tfrac12+\tfrac12=1$.

Answer 2

Is it permitted to use Taylor expansion? Like

$$ \arcsin(x) = \int^x_0 \frac{1}{\sqrt{1-t^2}}\,dt = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \frac{35x^9}{1152} + \dotsc $$