Calculate the limit $\lim_{(x,y)\to (0,0)} \frac{x-\cos(x^2+y^2)-\arctan(x)+1}{x^2+y^2}$ Find the limit, $$\lim_{(x,y)\to (0,0)} \frac{x-\cos(x^2+y^2)-\arctan(x)+1}{x^2+y^2}.$$
My attempt:
I have tried several paths for evaluating the limit. For instance, $y=0$, $x=0$, $y=x$, $x=\tan y$, etc. I always obtain $0$. But I haven't been able to prove that the limit is equal to $0$.

You may use the following limits which you can easily show using Taylor or L'Hosp.:

*

*$\lim_{t\to 0}\frac{1-\cos t}{t^2}=\frac 12$, hence, $\lim_{t\to 0}\frac{1-\cos t}{t}=0$

*$\lim_{x\to 0}\frac{x-\arctan x}{x^2}=0$
Now, just rewrite the given expression as follows:
\begin{eqnarray*}\frac{x-\cos(x^2+y^2)-\arctan(x)+1}{x^2+y^2}
& = & \underbrace{\frac{1-\cos(x^2+y^2)}{x^2+y^2}}_{\stackrel{(x,y)\to (0,0)}{\longrightarrow}0} + \underbrace{\frac{x-\arctan x}{x^2+y^2}}_{\left|\frac{x-\arctan x}{x^2+y^2}\right|\leq\left|\frac{x-\arctan x}{x^2}\right|\stackrel{(x,y)\to (0,0)}{\longrightarrow}0} \\
\end{eqnarray*}
