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I am trying to prove that:

$$\lim_{x \to 0} \frac{1}{1+x} =1.$$

Here's how I started my proof,

$$|f(x)−1| = \left|\frac{1}{1+x} -1 \right| = \frac{x}{1+x} < \frac{\delta}{1+x}$$

I don't know how to continue now.

Arturo Magidin
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john grey
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1 Answers1

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If $|x|<\frac12$, then $1+x\in\left(\frac12,\frac32\right)$; in particular, $1+x>\frac12$, and therefore $\left|\frac x{1+x}\right|<2|x|$. So, take $\delta=\min\left\{\frac12,\frac\varepsilon2\right\}$, and then\begin{align}|x|<\delta&\implies\left|\frac x{1+x}\right|<2|x|\text{ (since $|x|<\frac12$)}\\&\implies\left|\frac x{1+x}\right|<\varepsilon\end{align}(since $|x|<\frac\varepsilon2$).

José Carlos Santos
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  • why do u assume that |x|< 1/2? it means that delta=1/2? – john grey Dec 21 '21 at 18:57
  • $x$ is tending to $0$. We can assume that $|x|$ is as small as we wish when computing the limit. – robjohn Dec 21 '21 at 18:58
  • No, it does not mean that $\delta=\frac12$. As I wrote, $\delta=\min\left{\frac12,\frac\varepsilon2\right}$. I started with $|x|<\frac12$ because then $\left|\frac x{1+x}\right|<2|x|$. And now it is quite easy to see which $\delta$ I should pick so that $|x|<\delta\implies\left|\frac x{1+x}\right|<\varepsilon$. – José Carlos Santos Dec 21 '21 at 19:00