How to give an Epsilon-Delta proof of this limit $ \lim_{x\to 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2} $?

Question

How to give an Epsilon Delta proof of the following limit:

$$ \lim_{x\to 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2} $$

Thanks for your help in advance!

Answer 1

observe $$ \dfrac{1-\sqrt{x}}{1-x} = \dfrac{1}{1+\sqrt{x}} $$ and you can now guess that the limit is 0.5 to prove find $$\begin{align*} \left| \dfrac{1}{1+\sqrt{x}} - \dfrac{1}{2} \right| &= \left| \dfrac{1-\sqrt{x}}{2(1+\sqrt{x})} \right|\\ &= \left| \dfrac{x-1}{2(1+\sqrt{x})^2} \right|\\ &< \dfrac{|x-1|}{2} \end{align*} $$ Now just take $\delta = 2\varepsilon$

Answer 2

1. Subtract $1/2$ from the function to get

$\dfrac{1-\sqrt x}{1-x}-\dfrac12=\dfrac{1-2\sqrt x+x}{2(1-x)}$

2. Apply $(a-b)^2=a^2-2ab+b^2$ and $(c+d)(c-d)=c^2-d^2$ (do not be afraid to put in square roots for $b$ and $d$) to obtain

$\dfrac{1-2\sqrt x+x}{2(1-x)}=\dfrac{1-\sqrt x}{2(1+\sqrt x)}=\dfrac{1-x}{2(1+\sqrt x)^2}$

3. Argue that if $|1-x|=\delta$ is sufficiently small, then the denominator in the last expression given above will be greater than $2$. Use this to find $\delta$ as a function of $\epsilon$ so that the absolute value of $\dfrac{1-\sqrt x}{1-x}-\dfrac12=\dfrac{1-x}{2(1+\sqrt x)^2}$ will be $<\epsilon$.

$\delta=2\epsilon$. You may find a somewhat larger $\delta$ is allowed if you can sharpen the bound on the denominator in Step 3.