Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$.

Question

Could someone please explain a step in the following proof?

Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$.

The above limit exists if for every $\varepsilon > 0$, there exists a real number $\delta > 0$ such that if $0<|x-1|<\delta$, then $\left |\frac{1}{x+1}-\frac{1}{2} \right |=\left | \frac{2-x-1}{2(x+1)} \right |=\left | \frac{1-x}{2(x+1)} \right |=\frac{|1-x|}{2|x+1|}=\frac{|x-1|}{2|x+1|}<\varepsilon$.

I understand that $|x-1|$ can be made small, but for $|x+1|$, why does the solution say the condition "if $|x-1| < 1$, $x+1\in (1,3)$"? Why and how is the condition $|x-1| <1$ made?

Answer 1

So far we have $$\left |\frac{1}{x+1}-\frac{1}{2} \right |=\left | \frac{2-x-1}{2(x+1)} \right |=\left | \frac{1-x}{2(x+1)} \right |=\frac{|1-x|}{2|x+1|}=\frac{|x-1|}{2|x+1|}<\varepsilon$$ In order to achieve the last inequality, we have to control the denominator as well as the numerator.

By assuming $|x-1|<1$ we can control $|x+1|$, therefore the fraction, $ \frac{|x-1|}{2|x+1|}$ could be made less than $\epsilon$ by choosing a small enough $ \delta$ .

Answer 2

The condition for $|x-1|<1$ is just a trick, you can set any number for convenience since we want to show that there exists delta.

First, let us observe the following definition of limit $$\forall \varepsilon > 0, \exists \delta > 0 \ni 0 < |x-1| < \delta \implies \bigg|\frac{1}{x+1}-\frac{1}{2}\bigg|< \varepsilon$$

So, our goal is to show that $|x-1|<\delta \implies \bigg|\frac{1}{x+1}-\frac{1}{2}\bigg|< \varepsilon$ To do this, we have to realise that limit is a concept of getting closer and closer to a certain value. Hence, we want to "set a limit" for the value of $x$ which we will observe. In this case, we want $x$ to get closer and closer to $1$. Hence, we just simply restrict $|x-1|<1$. This can be any convenient number. You can even set to $1/2$ if you want to.

Now, moving on observe that $\bigg|\frac{1}{x+1} - \frac{1}{2}\bigg| = \bigg|\frac{2-x-1}{2(x+1)}\bigg|= \frac{1}{2}\bigg|\frac{x-1}{x+1}\bigg|$. Since we restrict $|x-1|<1$, we have $1<x+1<2$ and thus $1<|x+1|<2$. As a consequence we have $\bigg|\frac{1}{x+1} - \frac{1}{2}\bigg|<\frac{1}{2}\frac{|x-1|}{1}=\frac{1}{2}|x-1|$.

Finally, we can prove the limit easily using the reasoning above by fixing $\varepsilon > 0$ and set $\delta = \min\{\frac{\varepsilon}{2},1\}$. Observe that : $$\bigg|\frac{1}{x+1} - \frac{1}{2}\bigg| = \frac{1}{2}\bigg|\frac{x-1}{x+1}\bigg|< \frac{1}{2}|x-1|<\frac{1}{2}\frac{\varepsilon}{2}<\varepsilon$$ Thus, we complete the proof.