$$ \text{Evaluate the following limit: }
A= \lim_{{(x,y) \to (0,0)}} \frac{{x^4 + y^4}}{{\sqrt{{x^2 + y^4}}}}$$
I've tried using polar coordinates, but am having trouble solving this.
$$
\lim_{{r \to 0}} \frac{{(r\cos(\theta))^4 + (r\sin(\theta))^4}}{{\sqrt{{(r\cos(\theta))^2 + (r\sin(\theta))^4}}}}
$$
After substituting, my first thought was to rationalize the denominator, which resulted in this:
$$
\lim_{{r \to 0}} \frac{{\sqrt{{(r\cos(\theta))^2 + (r\sin(\theta))^4)}}{{(r\cos(\theta))^4 + (r\sin(\theta))^4)
}}}}{{(r\cos(\theta))^2 + (r\sin(\theta))^4}}
$$
But from here I'm having trouble moving forward.
Can anyone provide any hints or tips? Would be greatly appreciated!
Best Answer
$$\lim_{{(x,y) \to (0,0)}} \frac{{x^4 + y^4}}{{\sqrt{{x^2 + y^4}}}}$$ Note that : $$0\le\left| \frac{{x^4 + y^4}}{{\sqrt{{x^2 + y^4}}}}\right| \le \left| \frac{{x^4 }}{{\sqrt{{x^2+y^4}}}}\right|+\left| \frac{{y^4 }}{{\sqrt{{x^2 + y^4}}}}\right|\le\left| \frac{{x^4 }}{{\sqrt{{x^2}}}}\right|+\left| \frac{{y^4 }}{{\sqrt{{ y^4}}}}\right|=|x^3|+y^2$$ and the last term approaches $0$ as $(x,y)\to (0,0)$. You conclude applying the squeeze theorem.