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I want to prove that the function $f(x)=\frac{x}{1+x}$ is continuous for $x\geq 0$. I'm using the definition of continuity taking $\varepsilon > 0\:\:and\:\:\delta=\varepsilon\cdot(1+x)(1+c)$:

$$\left | f(x)-f(c)\right |=\left | \frac{x}{1+x}-\frac{c}{1+c} \right |=\left | \frac{x-c}{(1+x)\cdot (1+c)}\right |= \frac{\delta }{(1+x)\cdot (1+c)} = \frac{\varepsilon \cdot (1+x)\cdot (1+c)}{(1+x)\cdot (1+c)}=\varepsilon $$ Does this prove that $f(x)$ is continuous? Thanks

EDIT: $$\left | f(x)-f(c)\right |=\left | \frac{x}{1+x}-\frac{c}{1+c} \right |=\left | \frac{x-c}{(1+x)\cdot (1+c)}\right |=\left | \frac{1}{(1+x)(1+c)} \right |\cdot \left | x-c \right |\leq 1\cdot \left | x-c \right |\leq \delta =\varepsilon $$

JaviBT
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1 Answers1

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To prove that $f$ is continuous in $c$ you have to show that for an arbitrary $\varepsilon>0$ there exists a $\delta:=\delta(\varepsilon,c)$ such that $$ |f(x)-f(c)|\le \varepsilon,\; \forall x : |x-c|\le \delta$$

With $\delta:=\delta(\varepsilon,c)$ I want to express that $\delta$ may depend on $c$ and $\varepsilon$. It must not depend on $x$. You define delta as $$\varepsilon\cdot(1+x)(1+c)$$ which apparently depends on $x$. So this $\delta$ does not fulfill the requirements. But you have

$$\left | f(x)-f(c)\right |=\left | \frac{x}{1+x}-\frac{c}{1+c} \right |=\left | \frac{x-c}{(1+x)\cdot (1+c)}\right |\lt \frac{\delta }{(1+x)\cdot (1+c)} \le \delta=\varepsilon$$ because $x\ge0$ and $c\ge 0$ and therefore $(1+x)\cdot(1+c)\ge 1$.

miracle173
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