Use the definition of limit to establish $ \lim_{x\rightarrow 2} \frac{1}{1-x}= -1$

Question

Let $\epsilon > 0$ . $|\frac{1}{1-x} -(-1)|= |\frac{2-x}{1-x}| < \epsilon$

Would my next step be $\frac{-1}{(1-x)} |x-2|$

Answer 1

You need to show that $\left|\frac{2-x}{1-x}\right|<\epsilon$ provided $0<|x-2|<\delta$, and you need to determine what that value $\delta$ is.

Notice that $\left|\frac{2-x}{1-x}\right|=\frac{|2-x|}{|1-x|}$. So if $0<|x-2|<\delta$, then the numerator is $<\delta$. How do we control the denominator? The only problem happens if $x$ is too close to $1$ where the denominator is zero. So we have to choose $\delta$ so that $x$ stays away from $0$. If we choose $\delta \leq \frac{1}{2}$, then whenever $0<|x-2|<\delta$, i.e., $x \in (2-\delta,2+\delta)$, we know that $x>2-\delta \geq 2-\frac{1}{2}=\frac{3}{2}$. Hence $x-1 \geq \frac{3}{2}-1=\frac{1}{2}$. This implies $\frac{1}{|x-1|} \leq \frac{1}{1/2}=2$.

Combining these, we have $\frac{|2-x|}{|1-x|} \leq 2\delta$. Now it should be clear how to choose $\delta$ in order to make this $\leq \epsilon$ (keeping in mind that we also need $\delta \leq \frac{1}{2}$).