Calculus - prove that limit doesn't exists using esplion and delta

Question

Using epsilon and delta proof this :

$$\lim_{x \to 3} \sqrt{x+1}\neq1$$

Exist $\varepsilon>0$ All $\delta>0$ so for some $x$ that appiles $0<|x-3|<\delta$ and however $|\sqrt{x+1}-1|\geq\varepsilon$

I don't really know how to approach this, any help will be appreciated.

Answer 1

Choose any $\varepsilon < 1$. For whatever $\delta$ you can choose $x = \min\lbrace 4, 3+\frac{\delta}{2}\rbrace$ So:

$$\vert \sqrt{x+1}-1\vert > \vert \sqrt{3+1}-1\vert = 1 > \varepsilon$$