### Quiz 2.3:

Question:

Prove$$\lim \limits _{x \rightarrow 4} \sqrt{x}=2$$

Solution:

Given $$\epsilon >0$$, we want to find $$\delta$$ , such that $$0<|x-4|< \delta \implies |\sqrt{x}-2|< \epsilon$$.
So for $$0<|x-4|< \delta$$
$$- \delta < x-4 < \delta , x \neq 4$$
$$4-\delta < x < 4+ \delta$$

And for $$|\sqrt{x}-2| < \epsilon$$
$$- \epsilon < \sqrt{x}-2 < \epsilon$$
$$2- \epsilon < \sqrt{x} < 2 + \epsilon$$
$$(2- \epsilon )^2 < x < (2+ \epsilon)^2$$
So we need to find $$\delta$$ such that
$$(2- \epsilon)^2< 4 - \delta < x < 4+ \delta < (2+ \epsilon)^2$$
So we need $$(2- \epsilon)^2 < 4 -\delta$$ and $$4+ \delta < (2+ \epsilon)^2$$
$$\delta < 4-(2- \epsilon)^2$$ and $$\delta< (2+ \epsilon)^2 -4$$
So we need the smaller $$\delta$$, so we take $$\delta < 4-(2- \epsilon )^2$$