I was a given a problem to solve (if the limit exists):
$$ \lim_{(x,y)\to(0,0)} \cos\left(\frac{x^3-y^3}{x^2+y^2}\right) .$$
My Approach:
Approaching along the path $y=0$, yield the limit to be $1$. Similarly, going along the path $y=mx$ yields the limit to be $1$. Thus, it seems that if the limit does exist, the limiting value has to be $1$.
To show that it exists, I decided to use the $\epsilon – \delta$ approach (I am unable to think of an algebraic approach).
It will be sufficient to check whether the limit:
$$ \lim_{(x,y)\to(0,0)} \frac{x^3-y^3}{x^2+y^2} $$ exists or not and if it does exist it should be equal to $0$.
We know that for some $\epsilon \in \mathbb{R}$, $$ 0< \left|\frac{x^3-y^3}{x^2+y^2}\right| < \epsilon,$$ we must be able to choose a $\delta$ such that $0<\sqrt{x^2+y^2} < \delta$ for the limit to exist.
Now,
$$ \left|\frac{x^3-y^3}{x^2+y^2}\right| = \left| \frac{(x-y)(x^2+y^2+xy)}{x^2+y^2} \right| < \frac{|(x-y)||xy|}{x^2+y^2} \leq \frac{|(x-y)|}{2} < |x-y| = \sqrt{x^2+y^2-2xy}.$$
We have 2 cases: Case I: $xy>0$ This will result in ,
$$ \left|\frac{x^3-y^3}{x^2+y^2}\right| < \sqrt{x^2+y^2}< \epsilon$$ Here $\delta$ can be chosen as $\epsilon$.
Case II: $xy<0$. This is where I am not able to proceed further.
Is there any alternative approach? Or can the case II part be proven?
Best Answer
The limit exists and equals $1$. This follows from the squeeze theorem and the inequalities $$ 0\leq\frac{|x^3-y^3|}{x^2+y^2}\leq |x|+|y|. $$ Indeed, we have that $$ |x^3-y^3|\leq |x|^3+|y|^3\leq (|x|+|y|)(x^2+y^2), $$ where the second inequality follows by expanding out the terms on the right and observing that the extra two cross terms are non-negative.