# Limit only using squeeze theorem $\lim_{(x,y)\to(0,2)} x\,\arctan\left(\frac{1}{y-2}\right)$

Someone can help me to solve this limit only using squeeze theorem

$$\lim_{(x,y)\to(0,2)} x\arctan\left(\frac{1}{y-2}\right)$$

I have done the substitution: $$\lim_{(x,v)\to(0,0)} x\arctan\left(\frac{1}{v}\right)$$

$$\arctan\left(\frac{1}{v}\right) \sim \frac{1}{v}$$

$$\lim_{(x,v)\to(0,0)} \frac{x}{v}$$

$$0\le \frac{|x|}{|v|}\le$$

Now I'm stucked, I don't know how to continue the inequality. Someone can help me?

• Your substitution is only correct near 0, not near $\infty$. – Miguel Jan 27 '16 at 22:58

Note that if $y\ne 2$ then $$0\le \left|x\arctan\left(\frac{1}{y-2}\right)\right| \le \frac{\pi}{2}|x|.$$ Now squeeze.