Answer
(a) We show that
$|{x^3}| \le |x|\left( {{x^2} + {y^2}} \right)$, ${\ \ }$ $|{y^3}| \le |y|\left( {{x^2} + {y^2}} \right)$
(b) We show that
$|f\left( {x,y} \right)| \le |x| + |y|$
(c) By Squeeze Theorem we obtain the limit:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = 0$
Work Step by Step
(a) Since ${y^2} \ge 0$, so
${x^2} \le {x^2} + {y^2}$
For $x>0$,
$|x|{x^2} \le |x|\left( {{x^2} + {y^2}} \right)$
Since ${x^2} = |{x^2}|$,
$|{x^3}| \le |x|\left( {{x^2} + {y^2}} \right)$
For $x<0$,
$ - |x|{x^2} \ge - |x|\left( {{x^2} + {y^2}} \right)$
Since ${x^2} = |{x^2}|$, we may write the last equation as
$ - |{x^3}| \le - |x|\left( {{x^2} + {y^2}} \right)$
Multiply both sides by $ - 1$ gives
$|{x^3}| \le |x|\left( {{x^2} + {y^2}} \right)$
Hence, for any $x$ we obtain $|{x^3}| \le |x|\left( {{x^2} + {y^2}} \right)$.
Similarly, $|{y^3}| \le |y|\left( {{x^2} + {y^2}} \right)$.
Thus,
$|{x^3}| \le |x|\left( {{x^2} + {y^2}} \right)$, ${\ \ }$ $|{y^3}| \le |y|\left( {{x^2} + {y^2}} \right)$
(b) We have $f\left( {x,y} \right) = \frac{{{x^3} + {y^3}}}{{{x^2} + {y^2}}}$. Taking the absolute value of both sides gives
$|f\left( {x,y} \right)| = |\frac{{{x^3} + {y^3}}}{{{x^2} + {y^2}}}| = \frac{{|{x^3} + {y^3}|}}{{|{x^2} + {y^2}|}}$
By triangle inequality we have
$|{x^3} + {y^3}| \le |{x^3}| + |{y^3}|$
and also
$|{x^2} + {y^2}| \le |{x^2}| + |{y^2}| = {x^2} + {y^2}$
Thus,
$|f\left( {x,y} \right)| \le \frac{{|{x^3}| + |{y^3}|}}{{{x^2} + {y^2}}}$
From part (a) we obtain
$|{x^3}| \le |x|\left( {{x^2} + {y^2}} \right)$, ${\ \ }$ $|{y^3}| \le |y|\left( {{x^2} + {y^2}} \right)$
Thus,
$|f\left( {x,y} \right)| \le \frac{{\left( {|x| + |y|} \right)\left( {{x^2} + {y^2}} \right)}}{{{x^2} + {y^2}}}$
Hence, $|f\left( {x,y} \right)| \le |x| + |y|$.
(c) From part (b) we obtain $|f\left( {x,y} \right)| \le |x| + |y|$.
We may write
$0 \le |f\left( {x,y} \right)| \le |x| + |y|$
Taking the limit, we obtain
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} 0 \le \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) \le \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} |x| + |y|$
The two limits on both ends are equal to $0$.
Therefore, by Squeeze Theorem we obtain the limit:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = 0$
