# Use epsilon-delta definition to show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$.

Could someone check whether my solution is okay? The textbook chose $\delta = \operatorname{min}(1,2\epsilon)$.

Use epsilon-delta definition to show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$.

Let $\epsilon >0$. Then $|\frac{1}{x+1}-\frac{1}{2}|<\epsilon$ is true if $|\frac{1}{x+1}-\frac{1}{2}|=\frac{|x-1|}{2|x+1|}<\epsilon$.

If $\delta \leq 1$, then $0<|x-1|<1$ or $-1<x-1<1$ or $0<x<2$. Then $x+1\in [1,3]$. Then $x+1<3$.

Then $\frac{|x-1|}{2|x+1|}<\frac{\delta}{(2)(3)}<\epsilon$ if $\delta = \operatorname{min}(1,6\epsilon)$.

• If $1<x+1<3\Rightarrow \frac{1}{|x+1|}=\frac{1}{x+1}>\frac{1}{3}$. But you are almost there, just use that $1<x+1\Rightarrow \frac{1}{|x+1|}=\frac{1}{x+1}<1$. – PtF Jun 3 '18 at 1:49
• If you want $\frac{1}{x+1} < N$ then you need to show that $x+1 > N > 0$. – steven gregory Jun 3 '18 at 2:01
• @PtF Thank you for taking the time in explaining. I got the right answer now. :) – numericalorange Jun 3 '18 at 2:06

Using $\delta = \operatorname{min}(1,2\epsilon)$
\begin{align*} 0<|x-1|<\delta \leq 1 &\implies -1 < x-1 < 1 \\ &\implies 1 < x+1 < 3 \\ &\implies 2 < 2(x+1) < 6 \\ &\implies \frac{1}{2} > \frac{1}{2(x+1)} > \frac{1}{6} \\ &\implies \frac{1}{6} < \frac{1}{2(x+1)} < \frac{1}{2} &\text{as $-\frac{1}{2}<\frac{1}{6}$}\\ &\implies \left| \frac{1}{2(x+1)} \right| < \frac{1}{2} \\ \end{align*}
Then we get \begin{align*} \left| \frac{1}{x+1} - \frac{1}{2} \right| &= \left|\frac{1}{2(x+1)}\right||x-1| \\ &< \frac{1}{2}|x-1| \\ &< \frac{1}{2}(2\epsilon) & \text{since $|x-1|<\delta \leq 2\epsilon$}\\ &= \epsilon \end{align*}