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I Need to prove by $\varepsilon-\delta$ definition that:

$$\lim_{x\to0}\frac{1}{x+1} = 1$$

So far I have written

Let $\epsilon>0$.

Want to find $\delta>0$ such that $|f(x)-1|<\epsilon$ if $0<|x|<δ$

We have $|f(x)-1| = |[1/(x+1)]-[(x+1)/(x+1)]| = |-(x)/(x+1)| = |x/(x+1)|$

Thus $|x/(x+1)|<\epsilon$

I need to get this in so $|x|<$ whatever, but I'm not sure how to do so without having $x$ on both sides which I'm pretty sure isn't allowed. How can I manipulate this to get it in the right form?

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Suppose first that $|x|<1/2$. Then, $$ -\frac{1}{2}<x<\frac{1}{2}\implies \frac{1}{2}<x+1<\frac{3}{2}\implies \frac{1}{2}<|x+1|<\frac{3}{2}\implies\frac{2}{3}<\frac{1}{|x+1|}<2 $$ Thus, under the assumption that $|x|<1/2$, we have $$ \frac{|x|}{|x+1|}\leq 2|x|. $$ If, in addition, we assume that $|x|<\epsilon/2$, then it follows that $$ \frac{|x|}{|x+1|}\leq 2|x|<2\frac{\epsilon}{2}=\epsilon, $$ which is what you want. This tells you that you could choose $\delta=\min\{1/2, \epsilon/2\}$.

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Hint

as you are near $0$,

you can add the condition

$$ \frac{-1}{2}<x< \frac{ 1}{2 }$$ to get

$$|\frac{x } {x+1}|<2|x|$$

and take $$\eta=\min(\frac{1}{ 2},\frac{\epsilon}{ 2})$$

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Choose $|x|<\frac{1}{2}$ then, $$ \frac{1}{2}<1-|x|<|x+1|$$ which means that $$\left|\frac{x}{x+1}\right|<2|x|<2\varepsilon$$ Hence takes $\delta= min(\frac{1}{2}, \frac{\varepsilon}{2})$

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