I'm working on showing that $$\lim \limits_{x \to 3} \sqrt{x+1}=2$$ I've made $\delta ≤4$ to ensure that $\sqrt{x+1}≥0$, but I'm stuck on how to proceed from here. I've tried messing with numerical bounds, but I can't seem to find the connection between $\lvert x-3 \rvert$ and $\lvert\sqrt{x+1}-2\rvert$.

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    $\begingroup$ Try multiplying by the conjugate. $\endgroup$ – David Bowman Aug 22 '17 at 1:46
  • $\begingroup$ So with the conjugate I can let $\delta =$ min{$\epsilon, 4$}? $\endgroup$ – Simplicio Aug 22 '17 at 1:49



If $|x-3|<4$, we have $$-1<x<7, $$$$0< \sqrt{x+1} < \sqrt{8}$$

$$2 < \sqrt{x+1}+2 < 2+ 2\sqrt{2}$$

$$\frac{1}{\sqrt{x+1}+2} < \frac12$$

$$|\sqrt{x+1}-2| \leq \frac{|x-3|}2$$


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