Let $f(x,y,z)=\frac{x}{y-z}$. $\lim\limits_{{x,y,z} \to {2,2,1}} \frac{x}{y-z}=2$. If $\epsilon=.5$, find $\delta$.
Here's (grossely) what I have done up to now:
$y \in [2-\delta, 2+\delta]$ and $z \in [1- \delta, 1+ \delta]$ $\implies ((2-\delta)-(1-\delta))\le(y-z)\le((2+\delta)-(1-\delta))$
$\implies 1\le(y-z)\le(2\delta+1)$
We know $\lvert\frac{x}{y-z}-2\rvert\le0.5$
$\lvert\frac{x}{y-z}-2\rvert=\lvert\frac{(x-2(y-z))}{y-z}\rvert\le\lvert\frac{x-2(2\delta+1)}{1}\rvert=\lvert\ (x-2)-4\delta\rvert\le|\delta-4\delta|=3\delta$
So $\delta=3\epsilon=1.5$. However, my answer doesn't appear to be correct. Where did I go wrong? What am I missing?
Some other problems:
$\lvert\frac{(x-2(y-z))}{y-z}\rvert\le\lvert\frac{x-2(2\delta+1)}{1}\rvert$ is not true. The change in denominator is justifiable (after making the necessary correction from the error I've already pointed out), but note that you have made the numerator substitution as large as possible, even though it is being subtracted from $x$.
$\lvert\ (x-2)-4\delta\rvert\le|\delta-4\delta|$ is not true. $|a| \le |c|$ does not in general imply that $|a+b| \le |c+b|$. However, this is easily rectifiable, because $\lvert\ (x-2)-4\delta\rvert\le\lvert\ (x-2)\rvert +\lvert4\delta\rvert\le5\delta$.
$\lvert\frac{x}{y-z}-2\rvert < 3\delta \implies \delta = 3\epsilon$ You've got that backwards. Remember: we need $\lvert\frac{x}{y-z}-2\rvert < \epsilon$, so what you need here would be $\epsilon = 3\delta$ (if $3\delta$ had been correct in the first place.)