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Prove $\lim_{x \to 1} \frac{1}{1+x} = \frac{1}{2}$ with $ \varepsilon - \delta $ definition

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$\lim\limits_{x \to 1} \frac{1}{1+x} = \frac{1}{2}$

Here is my attempt

Prep work:

$$\Bigl| \frac{1}{1+x} - \frac{1}{2} \Bigr| = \frac{1}{2} \Bigl| \frac{1-x}{1+x}\Bigr|=\frac{1}{2}\cdot |x-1| \cdot \frac{1}{|1+x|}$$

If we assume $|x-1|< 1$ is true, it would follow $$ -1<x-1 \implies 1<x+1 \implies \frac{1}{|1+x|}<1$$ then we'd have $$\frac{1}{2}\cdot |x-1| \cdot \frac{1}{|1+x|} < \frac{1}{2}\cdot |x-1| \cdot 1 \quad (*)$$

This suggests a choice of $\delta = 2 \varepsilon$

Proof.

Let $\varepsilon > 0$ and set $\delta = \min \{1, 2 \varepsilon \}$.

We assume $0<|x-1|<\delta$ which means $|x-1|<1 \land |x-1|<2 \varepsilon$

Then, by $(*)$ and $|x-1|<1$ indeed we have $$\frac{1}{2}\cdot |x-1| \cdot \frac{1}{|1+x|} < \frac{1}{2}\cdot |x-1| \cdot 1 $$

and by $|x-1|<2 \varepsilon$$$\frac{1}{2}\cdot |x-1| < \frac{1}{2} \cdot 2 \varepsilon=\varepsilon$$

End of proof.

Is this correct, is it properly structured? thanks


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