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This is what I've done: $$\left|\frac{x}{1+x} - \frac{1}{1+1}\right| = \left|\frac{x-1}{2+2x}\right| = \left|x-1\right|\frac{1}{\left|2+2x\right|} < \epsilon $$

How should I find the lower bound $\left|2+2x\right|$? Do I need to introduce another variable?

SchrodingersCat
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lsy
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2 Answers2

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You can assume that $x \in (0,2)$, say, so that $|2+2x| > 2$. Therefore $\frac1{|2+2x|} < \frac12$.

Théophile
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  • Why is it that we can just take x∈(0,2). Don't we need to consider the entire domain? – lsy Nov 13 '15 at 18:56
  • @shiy, can you write down the definition of continuity? It explicitly wants that you find an open interval (dependent on $\epsilon$) such that $|f(x)-f(1)| <\epsilon$ for $x$ inside that interval, not the entire domain. – Ennar Nov 13 '15 at 19:49
  • @shiy We don't need to consider the entire domain, because we're only interested in continuity at one point. We can look at the neighbourhood immediately around $x=1$. I chose $(0,2)$ for the sake of convenience, but if you really wanted to, you could assume that $x \in (0.9999,1.0001)$, for example. The one thing that should be checked is whether the function is actually defined on the given interval; in the case of $(0,2)$, we're safe, because the only point where $f$ is undefined is $x=-1$. (And the other interval I mentioned is similarly safe.) – Théophile Nov 13 '15 at 19:50
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    Ah ok I think I finally got what the definition means. Thank you so much! – lsy Nov 13 '15 at 20:12
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It is a composition of continuous functions. Division function only have a singularity in 0, in our case, in x+1=0, but our case is x=1, then x+1=2. Then, a composition of continuous functions is continuous.

Raúl Aparicio Bustillo
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