epsilon delta limit

Question

By using the definition of limit to show that

$\lim\limits_{(x,y,z)\to (0,0,0)}\frac{x(y^2 + z^2)}{x^2+y^2+z^2}=0$

I dun know how to prove it, could anyone can give me some idea? Thank you so much

Answer 1

hist: use the inequality $$|\frac{x(y^2+z^2)}{x^2+y^2+z^2}| \leq |\frac{x(x^2+y^2+z^2)}{x^2+y^2+z^2}| = |x|$$

Answer 2

Hint:

$\left | \frac{x(y^2 + z^2)}{x^2 + y^2 + z^2} \right| \leq | x |$

Answer 3

$$\left|\frac{x\left(y^{2}+z^{2}\right)}{x^{2}+y^{2}+z^{2}}\right|=\frac{\left|x\right|\left(y^{2}+z^{2}\right)}{x^{2}+y^{2}+z^{2}}=\frac{\left|x\right|y^{2}}{x^{2}+y^{2}+z^{2}}+\frac{\left|x\right|z^{2}}{x^{2}+y^{2}+z^{2}}\leq\frac{\left|x\right|y^{2}}{y^{2}}+\frac{\left|x\right|z^{2}}{z^{2}}=2\left|x\right|\longrightarrow0$$ when $x\rightarrow0$ . Tell me if something is not clear for you.

Answer 4

Answering questions like this often take some ingenuity with inequalities.

Notice that $$\left| \frac{x(y^2+z^2)}{x^2+y^2+z^2} \right| \le |x| \frac{x^2+y^2+z^2}{x^2+y^2+z^2}=|x|$$ since we are only adding a positive term to the numerator.

Now if we let $\epsilon > 0$, we need to show that there is a $\delta >0$ that when $\sqrt{x^2+y^2+z^2} < \delta$ we have our original quantity less than $\epsilon$. $\delta = \epsilon$ should be a good choice of $\delta$ in this case.

Answer 5

Alternatively, using spherical polars with the convention $x = r\cos\theta, y = r\sin\theta\sin\phi, z = r\sin\theta\cos\phi$, the quotient

$$\frac{x(y^2 + z^2)}{x^2+y^2+z^2} = r \cos\theta(\sin^2\theta(\sin^2\phi + \cos^2\phi)) = r\cos\theta\sin^2\theta$$

You can see now why I chose that particular version of polar coordinates: it enabled us to use $\sin^2 + \cos^2 = 1$ to simply the numerator nicely.

Thus we can now bound the expression as

$$\left| \frac{x(y^2 + z^2)}{x^2+y^2+z^2} \right| \leq | r\cos\theta\sin^2\theta | \leq r$$

Hence given $\epsilon > 0$, choose $\delta = \epsilon$. Then

$$ |(x,y,z) - (0,0,0)| = r < \delta \ \Rightarrow \left| \frac{x(y^2 + z^2)}{x^2+y^2+z^2} - 0 \right| < \epsilon$$

That is

$$\lim_{(x,y,z)\to(0,0,0)} \frac{x(y^2 + z^2)}{x^2+y^2+z^2} = 0$$