I want to give an epsilon delta proof that$$\lim_{x\to 3-}\sqrt{3-x}=0$$
This means I want to show that for every $\epsilon>0$ there exists a $\delta>0$ such that$$3-\delta<\left|x-3\right|<3$$ implies that $$\left|\sqrt{3-x}\right|<\epsilon$$
Usually my rough work would start of by doing something like $$\left|\sqrt{3-x}\right|=\frac{\left|3-x\right|}{\sqrt{3-x}}$$
and then say that's less than something else. I can't really do that in this case though because the denominator gets infinitely small. How can I get past this problem. Thanks.
Actually, since it is a left-hand limit, what you have to show is that for all $\epsilon > 0$, there is a $\delta > 0$ such that $3 - \delta < x < 3$ implies $|\sqrt{3-x}| < \epsilon$. Note that since $\sqrt{3-x} \geq 0$ for all $x \leq 3$ we have $|\sqrt{3-x}| = \sqrt{3-x}$. So, let's drop the absolute value.
Let $\epsilon > 0$. To figure out what to choose for $\delta$ let's square both sides of $\sqrt{3-x} < \epsilon$ getting $3-x < \epsilon^2$. This suggests letting $\delta = \epsilon^2$. With that choice of $\delta$ assume
$$3-\delta < x < 3.$$
This implies that $3 - \epsilon^2 < x$ and so $3 - x < \epsilon^2$. Taking the nonnegative square root of each side (which is ok on the left because $x < 3$) we get $\sqrt{3-x} < \epsilon$ as desired.