Prove that $\sqrt{x} + x$ approaches $2$ as $x$ approaches $1$

Question

How to prove using the epsilon-delta definition of a limit that $\sqrt{x} + x$ approaches $2$ as $x$ approaches $1$? What is the value of delta?

I'm having a difficulty of this one because of the equation $f(x) = \sqrt{x} + x$. Any help will do!

Answer 1

We have: $|\sqrt{x}+x - 2|= |\sqrt{x}-1+x-1|\le |\sqrt{x}-1|+|x-1|= \dfrac{|x-1|}{\sqrt{x}+1}+|x-1|\le 2|x-1|$. Thus for every $\epsilon > 0$, choose $\delta = \frac{\epsilon}{2}$, then if $0 < |x-1| < \delta \implies |\sqrt{x}+x-2| < 2\cdot\frac{\epsilon}{2}=\epsilon. $