How to show that this function is continuous I'm trying to show continuity of the function $$\frac {\ln(1+x^2+y^2)}{x^2+y^2}$$ for $(x,y)\neq 0$,
$$f(x,y)=1$$ for $(x,y)= 0$,
on $\mathbb{R}^2$
But I am not able to. The numerator is stuck for me as I don't know how to get around the log function. How can I simplify this expression or better get rid of the log?
Sorry for the sloppy edits
 A: HINT:
The function has a removable discontinuity at $(0,0)$ since $\lim_{(x,y)\to(0,0)}\frac{\log (1+x^2+y^2)}{x^2+y^2}=\lim_{z\to 0}\frac{\log (1+z)}{z}=1$.  
For all $x^2+y^2\ne 0$, view $x^2+y^2$ as a single variable and show continuity of $f(z)=\frac{\log (1+z)}{z}$ for $z\ne 0$
A: Hint:
$$
\lim_{(x,y)\rightarrow (0,0)}\frac {\ln(1+x^2+y^2)}{x^2+y^2}=1
$$
A: Hint:
$$\lim _{ x\rightarrow 0\\ y\rightarrow 0 }{ \frac { \ln { \left( 1+{ x }^{ 2 }+{ y }^{ 2 } \right)  }  }{ { x }^{ 2 }+{ y }^{ 2 } }  } =\lim _{ x\rightarrow 0\\ y\rightarrow 0 }{ \ln { \left( 1+{ x }^{ 2 }+{ y }^{ 2 } \right) ^{ \frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } }  } }  } =\ln { e=1 } $$
A: The function $f(x,y)=\dfrac{\ln(1+x^2+y^2)}{x^2+y^2}$ is defined and continuous on $\mathbf R^2\smallsetminus\{(0,0)\}$, as a composition and quotient of continous functions (polynomials and logarithms are continuous functions).
We can define a continuous continuation of $f$ at $(0,0)$ because $f$ has a limit at $(0,0)$. To see this, use polar coordinates: set $x=r\cos \theta,\ y=r\sin\theta$. Then for $(x,y)\neq (0,0)$,
$$f(x,y)=\frac{\ln(1+r^2)}{r^2}\xrightarrow[r\to0]{}1. $$
Thus we obtain a continuous fonction on $\mathbf R^2$ if we set
$$f(x)=\begin{cases}\dfrac{\ln(1+x^2+y^2)}{x^2+y^2}&\text{if}\enspace (x,y) \neq (0,0),\\1&\text{if}\enspace (x,y) = (0,0).\end{cases}$$
