$\lim\limits_{x \to +\infty} \frac{\cos(2x^{-\frac{1}{2}})-1}{\frac{\pi}{2}-\arctan(3x)}$

Question

I'm having difficulties with this limit. We haven't covered integrals, derivatives, or things such as L'Hopital's rule in our course yet. What is the correct process to go about this limit?

$$\lim\limits_{x \to +\infty} \frac{\cos(2x^{-\frac{1}{2}})-1}{\frac{\pi}{2}-\arctan(3x)}$$

Answer 1

We have that by $\frac{\pi}{2}-\arctan(3x)=\arctan \frac1{3x}$ and standard limits

$$\frac{\cos(2x^-\frac{1}{2})-1}{\frac{\pi}{2}-\arctan(3x)}=\frac{\cos(2x^-\frac{1}{2})-1}{4x^{-1}}\frac{\frac1{3x}}{\arctan \frac1{3x}}\cdot12\to -\frac12\cdot 1 \cdot12=-6$$

Answer 2

Set $\sqrt x=1/y$

and use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function? to find

$$\lim_{y\to0^+}\dfrac{\cos2y-1}{\arctan\dfrac{y^2}3}$$

$$=-\lim\dfrac{2\sin^2y}{\cdots}$$

$$=\cdots=-\dfrac2{\dfrac13}$$