i have to find the limit when $(x,y)\to (0,0)$ of $$\frac{\log(1+x^2 + y^2)}{x^2+y^2+yx^3}$$ According to wolframalpha the limit equals 1 but i have spent hours trying to solve it and i can't figure out how to get to the solution. I tried to use the sqeeze theorem but i couldn't find two bound functions approaching to 1, if this is the way could you give me some instructions or tips on how to find the two functions in theese cases? I've also tried using polar coordinates but got nowhere. Thanks in advance. Regards
Eliana
If $(x,y)\neq (0,0)$, then dividing the top and bottom of the fraction by $x^2+y^2$ we get:
$$\frac{\log(1+x^2+y^2)}{x^2+y^2+yx^3}=\frac{\frac{\log(1+x^2+y^2)}{x^2+y^2}}{1+\frac{yx^3}{x^2+y^2}}$$
Now, if $|y|<1$ and $|x|<1$, then $|yx^3|<|x^3|<|x|(x^2+y^2)$, so if not both $x,y$ are zero then $\left|\frac{yx^3}{x^2+y^2}\right|<|x|$. Therefore, $$\lim_{(x,y)\to (0,0)}\frac{yx^3}{x^2+y^2} = 0\hskip 1.5cm (1)$$ Also, substituting $x^2+y^2 = t$, we have $$\lim_{(x,y)\to (0,0)}\frac{\log(1+x^2+y^2)}{x^2+y^2}=\lim_{t\to 0}\frac{\log(1+t)}{t}=1\hskip 1.5cm (2)$$ From (1) and (2) we deduce that the limit in question is equal to $1$.