Proving a non-linear limit using epsilon and delta

Question

Prove the limit statement: $$\lim _{x\to 0}\left(\sqrt{4-x}\right)=2$$ My professor did similar problems in class but I can't seem to get this one right, any help appreciated.

Answer 1

EDIT: It appears I typed this all up and did the wrong limit. Still, this should be helpful, I think. I hope.

We want to show that $\forall$ $\epsilon \gt 0$ $ \exists \delta \gt 0$$ $such that $ \forall x$ with $|x - 0| = |x| \lt \delta, |\sqrt{4-x^2} -2| \lt \epsilon$.

Let's figure out how we need to choose $\delta$, then write up the formal proof.

$|\sqrt{4-x^2} -2| = |\sqrt{4-x^2} -2 * \frac {\sqrt{4-x^2} +2}{{\sqrt{4-x^2} +2}}| = |\frac{4 - x^ 2 - 4}{\sqrt{4-x^2}+2}| = |\frac{x^2}{{\sqrt{4-x^2}+2}}|$.

We want this expression to be less than $\epsilon$, so we have $|x^2| = |x|*|x| \lt \epsilon (\sqrt{4-x^2} +2)$, or, $|x| \lt \frac {\epsilon (\sqrt{4-x^2} +2)}{|x|}$.

Now we introduce the restriction $\delta = 1$ so that $-1 \lt x \lt 1$, and we see that under this restriction the right side of the inequality is minimized when the numerator is smallest and the denominator largest. Since $|x| \lt 1$, we see $|x| \lt \epsilon \frac{\sqrt{{4-1^2}} + 2}{1} = \epsilon (\sqrt{3} +2)$. Our choice of $\delta$ is now clear: $\delta = $min$\{1, \epsilon (\sqrt{3} +2)\}$.

Here's the formal write up:

Proof. Let $\epsilon \gt 0$ be given, choose $\delta = $min$\{1, \epsilon (\sqrt{3} +2)\} $, and suppose $0 \lt |x-0| = |x| \lt \delta$. Then $|f(x) - L| = |\sqrt{4-x^2} -2| = |\sqrt{4-x^2} -2 * \frac {\sqrt{4-x^2} +2}{{\sqrt{4-x^2} +2}}| = |\frac{4 - x^ 2 - 4}{\sqrt{4-x^2}+2}| = |\frac{x^2}{{\sqrt{4-x^2}+2}}| = \frac {|x|*|x|}{{{|\sqrt{4-x^2}+2|}}} \lt \frac{\epsilon(\sqrt{3} +2) * 1}{\sqrt{3} + 2} = \epsilon.$ Then, by the definition of the limit, as $x \to 0$, $\sqrt{4-x^2} \to 2$.

Answer 2

Hint: $$\sqrt{4-x}-2=\frac{(\sqrt{4-x})^2-2^2}{\sqrt{4-x}+2}=\frac{(\sqrt{4-x})^2-2^2}{\sqrt{4-x}+2}=\frac{|x|}{\sqrt{4-x}+2}$$