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prove using $\delta-\epsilon$ definition that $\lim_{x\to1} \frac{x}{1+x}=\frac{1}{2}$

$|\frac{x}{1+x} - \frac{1}{2}| = |\frac{-1}{2(1+x)}|=|\frac{1}{2(x-1)}|=|\frac{1}{2}||\frac{1}{x-1}|$

If $|x-1|<1$, then $1<1/|x-1|$. Here's where I got lost.

Basically I was thinking on making similar argument like Solving a problem using the definition of limit like Thomas Andrews and Surb answer, where we take $\delta$ to be minimum of 2 possible deltas. What I have is $|x-1|<\delta$ and $|f(x)-1/2|$ turns out to be $|\frac{1}{2}||\frac{1}{x-1}|$. I am not sure because what I have now is $\frac{1}{x-1}$ not $(x-1)$.

Thanks in advance.

aaaaaa
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  • you have a mistake, it should be $\left| \frac { x }{ 1+x } -\frac { 1 }{ 2 } \right| =\left| \frac { x-1 }{ 2(1+x) } \right| $ – haqnatural Nov 28 '17 at 20:00

2 Answers2

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Hint: $$\left| \frac { x }{ 1+x } -\frac { 1 }{ 2 } \right| =\left| \frac { x-1 }{ 2(1+x) } \right| <\left| \frac { x-1 }{ 2 } \right| <\left| x-1 \right| $$

haqnatural
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hint

$x $ is near $1$, so we can assume that

$|x-1|<\color {red}{1}$ or $0 <x <2$ which gives

$1 <x+1 <3$ and $\frac 13 <\frac {1}{x+1}<1$.

thus $$|\frac {x}{x+1}-\frac 12|=$$ $$|\frac {x-1}{2 (x+1)}|<\frac {|x-1|}{2} $$

so we just need to realise $$|x-1|<2\epsilon $$

We will take $$\delta=\min (\color {red}{1},2\epsilon) $$

hamam_Abdallah
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