# Prove that $\lim \limits_{x \to 0} {\sqrt{x+2}}=2$ (using Epsilon-Delta) [duplicate]

Can someone please guide me through the solving of $\lim \limits_{x \to 0} {\sqrt{x+2}}=2$ using the Epsilon-Delta definition?

I understand how to use $\epsilon-\delta$ in general cases and I know, so far, that:
Let $\epsilon > 0,$ $$\left|\sqrt{x+2}-2\right|<\epsilon\ whenever \left|x-0\right|<\delta$$ $$-\epsilon<\sqrt{x+2}-2<\epsilon\ whenever -\delta<x<\delta$$ Let's solve for $-\epsilon<\sqrt{x+2}-2<\epsilon\\2-\epsilon<\sqrt{x+2}<\epsilon+2\\(2-\epsilon)^2<{x+2}<(\epsilon+2)^2\\\epsilon^2-4\epsilon+2<x<\epsilon^2+4\epsilon+2$

And here I'm stuck. Is this even the right way to do it ?

## marked as duplicate by user296602, Community♦Mar 4 '18 at 20:56

• Should the limit be $\sqrt 2$? – sku Mar 4 '18 at 20:48
• You cannot prove this because it is not true. Maybe the limit is $\sqrt{2}$? – Emilio Novati Mar 4 '18 at 20:48
• The limit is sqrt (2) not 2. – hamam_Abdallah Mar 4 '18 at 20:52
• Oh that's why... I did not even have the reflex to check the limit directly. My professor made a mistake in the assignement, I'm sorry. – Prixeler Mar 4 '18 at 20:55

Note that

$$\lim \limits_{x \to 0} {\sqrt{x+2}}=\sqrt 2$$

thus following your way we should obtain

$$\epsilon^2-\sqrt 2\epsilon<x<\epsilon^2+\sqrt 2\epsilon$$

and

$$|x|<\min(\sqrt 2\epsilon-\epsilon^2,\sqrt 2\epsilon+\epsilon^2)=\sqrt 2\epsilon-\epsilon^2=\delta$$

$-\epsilon<\sqrt{x+2}-\sqrt2<\epsilon$

$\sqrt2-\epsilon<\sqrt{x+2}<\epsilon+\sqrt 2$

$(\sqrt 2-\epsilon)^2<{x+2}<(\epsilon+\sqrt 2)^2$

$\epsilon^2- 2\sqrt 2\epsilon < x< \epsilon^2+ 2\sqrt 2\epsilon$

Ignoring $\epsilon^2$ we get $- 2\sqrt 2\epsilon < x< 2\sqrt 2\epsilon$. Here $\delta = 2\sqrt 2 \epsilon$