# Proving a non-linear limit using epsilon and delta

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.

• Are you having trouble finding the $\delta$? – 3x89g2 Sep 13 '16 at 3:41
• Yeah, since it's a square root function, I can't seem to find the correct δ. – EconDude Sep 13 '16 at 3:44
• Tried squaring? – Zelos Malum Sep 13 '16 at 3:47
• My professor didn't mention squaring, how would I do that? – EconDude Sep 13 '16 at 3:48
• First of all, do you know what $\epsilon$ and $\delta$ are in this situation? What's the definition that you think you would need? – nbro Sep 13 '16 at 3:54

We want to show that $\forall$ $\epsilon \gt 0$ $\exists \delta \gt 0$$such that \forall x with |x - 0| = |x| \lt \delta, |\sqrt{4-x^2} -2| \lt \epsilon. Let's figure out how we need to choose \delta, then write up the formal proof. |\sqrt{4-x^2} -2| = |\sqrt{4-x^2} -2 * \frac {\sqrt{4-x^2} +2}{{\sqrt{4-x^2} +2}}| = |\frac{4 - x^ 2 - 4}{\sqrt{4-x^2}+2}| = |\frac{x^2}{{\sqrt{4-x^2}+2}}|. We want this expression to be less than \epsilon, so we have |x^2| = |x|*|x| \lt \epsilon (\sqrt{4-x^2} +2), or, |x| \lt \frac {\epsilon (\sqrt{4-x^2} +2)}{|x|}. Now we introduce the restriction \delta = 1 so that -1 \lt x \lt 1, and we see that under this restriction the right side of the inequality is minimized when the numerator is smallest and the denominator largest. Since |x| \lt 1, we see |x| \lt \epsilon \frac{\sqrt{{4-1^2}} + 2}{1} = \epsilon (\sqrt{3} +2). Our choice of \delta is now clear: \delta = min\{1, \epsilon (\sqrt{3} +2)\}. Here's the formal write up: Proof. Let \epsilon \gt 0 be given, choose \delta = min\{1, \epsilon (\sqrt{3} +2)\} , and suppose 0 \lt |x-0| = |x| \lt \delta. Then |f(x) - L| = |\sqrt{4-x^2} -2| = |\sqrt{4-x^2} -2 * \frac {\sqrt{4-x^2} +2}{{\sqrt{4-x^2} +2}}| = |\frac{4 - x^ 2 - 4}{\sqrt{4-x^2}+2}| = |\frac{x^2}{{\sqrt{4-x^2}+2}}| = \frac {|x|*|x|}{{{|\sqrt{4-x^2}+2|}}} \lt \frac{\epsilon(\sqrt{3} +2) * 1}{\sqrt{3} + 2} = \epsilon. Then, by the definition of the limit, as x \to 0, \sqrt{4-x^2} \to 2. Hint:$$\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}$\$