# Find $\lim\limits_{(x,y)\to(0,0)}\frac{e^{x^2+y^2}-x^2-y^2-1}{(x^2+y^2)^2}$

We have to calculate this limit of the multivariable function: $$\lim\limits_{(x,y)\to(0,0)}\frac{e^{x^2+y^2}-x^2-y^2-1}{(x^2+y^2)^2}$$ By setting $z=x^2+y^2$ then $z\to 0$ when $(x,y)\to(0,0)$, so if we apply de l'Hopital's rule twice, we find that the limit is equal to $\frac{1}{2}$. Now, I'm trying to calculate it by using the Squeeze Theorem (I've tried by using the ε-δ definition, without a result).

My thoughts so far:

I have tried to use the inequality $(x^2+y^2)^2\geq 2x^2y^2$ in order to overcome the problem that the denominator=0 and after some algebraic manipulation I can't get a result. Another thing I've tried is to use polar coordinates, but it doesn't seem to give an inequality.

What can I do?

Hint: Note that the integrand is radially symmetric, and so try putting things into polar coordinates by setting $r^2=x^2+y^2$. Then we are looking at the limit $$\lim_{r\rightarrow 0}\frac{e^{r^2}-r^2-1}{r^4}.$$ (Notice there is no $\theta$ term due to the symmetry of the integrand) To evaluate this, use the Taylor expansion $$e^x=1+x+\frac{x^2}{2}+O(x^3).$$
Using the exponential and geometric series, you get the quantitative bound $$\left|e^z-(1+z+\tfrac12 z^2)\right|\le \frac{|z|^3}{3!}·\frac{1}{1-\frac{|z|}{4}}.$$
Hint,use this $$1+x+\dfrac{x^2}{2}\le e^x\le 1+x+\dfrac{x^2}{2}+\dfrac{x^3}{2(3-x)},0\le x<3$$