Using triangle inequality to find $\lim_{(x,y)\to(0,0)}\frac{x^3-x^2y}{x^2+y^2+xy}$

Question

This is an exercise from my textbook where the problem is to find the limit of the function $\frac{x^3-x^2y}{x^2+y^2+xy}$ when $(x,y) \to (0,0)$.

So after changing to polar coordinates and simplifying I get the equivalent function $$r\cdot\frac{\cos^3(x)-\cos^2(x)\sin(x)}{1+\frac{1}{2}\sin(2x)}$$

In my textbook the limit of this function is solved by estimating the function using the triangle inequality (I am guessing thats $|x+y|\leq |x|+|y|$) :

$$\left|\frac{x^3-x^2y}{x^2+y^2+xy}\right|\leq r\cdot \frac{|\cos^3(x)|+\cos^2(x)|\sin(x)|}{1+\frac{1}{2}\sin(2x)} \leq r\cdot\frac{1+1}{1-\frac{1}{2}} $$

My question is how exactly was the triangle inequality used in this problem?

Answer 1

Because by the triangle inequality we obtain: $$\left|\frac{x^3-x^2y}{x^2+xy+y^2}\right|=\frac{x^2|x-y|}{x^2+xy+y^2}\leq\frac{x^2(|x|+|y|)}{x^2+xy+y^2}\leq$$ $$\leq\frac{x^2(|x|+|y|)}{x^2-|xy|+y^2}\leq\frac{x^2(|x|+|y|)}{\left(\frac{|x|+|y|}{2}\right)^2}=\frac{4x^2}{|x|+|y|}=4|x|\cdot\frac{|x|}{|x|+|y|}\leq4|x|\rightarrow0.$$