prove limit by definition

prove that $\displaystyle lim_{x\rightarrow2}\frac{x}{x+1}\neq1$ by epsilon delta

defintion.

we need to prove that their is $\displaystyle \varepsilon>0$ so for every $\displaystyle \delta>0$

their is x $\displaystyle |x-2|<\delta$ and $\displaystyle |\frac{x}{x+1}-1|\geq\varepsilon$

so we are free to choose our epsilon and our x

i choose $\displaystyle \varepsilon=2$

x cannot be equal to -1 because of the denominator.

and we need to choose x for which $\displaystyle |\frac{x}{x+1}-1|\geq2$

$\displaystyle |\frac{x-x-1}{x+1}|\geq2$ -> $\displaystyle |\frac{1}{x+1}|\geq2$ -> $\displaystyle \frac{1}{2}\geq|x+1|$

-> $\displaystyle -\frac{1}{2}\leq x+1\leq\frac{1}{2}$ -> $\displaystyle -\frac{3}{2}\leq x\leq-\frac{1}{2}$

but x=-1 there so we have a problem.

where is the problem in my way???

the prof solved it by $\displaystyle \varepsilon=1/6$

$\displaystyle 2>x>max\{2-\delta/2,1\}$

cant understand why he choose such x

Re: prove limit by definition

Quote:

Originally Posted by

**transgalactic** prove that $\displaystyle lim_{x\rightarrow2}\frac{x}{x+1}\neq1$ by epsilon delta

defintion.

we need to prove that their is $\displaystyle \varepsilon>0$ so for every $\displaystyle \delta>0$

their is x $\displaystyle |x-2|<\delta$ and $\displaystyle |\frac{x}{x+1}-1|\geq\varepsilon$

so we are free to choose our epsilon and our x

i choose $\displaystyle \varepsilon=2$

x cannot be equal to -1 because of the denominator.

and we need to choose x for which $\displaystyle |\frac{x}{x+1}-1|\geq2$

$\displaystyle |\frac{x-x-1}{x+1}|\geq2$ -> $\displaystyle |\frac{1}{x+1}|\geq2$ -> $\displaystyle \frac{1}{2}\geq|x+1|$

-> $\displaystyle -\frac{1}{2}\leq x+1\leq\frac{1}{2}$ -> $\displaystyle -\frac{3}{2}\leq x\leq-\frac{1}{2}$

but x=-1 there so we have a problem.

where is the problem in my way???

the prof solved it by $\displaystyle \varepsilon=1/6$

$\displaystyle 2>x>max\{2-\delta/2,1\}$

cant understand why he choose such x

Since the actual limit is 2/3, you will always be able to show that for x close enough to 2 that |x/(x+1)-1|>0.3, so you need to choose epsilon greater than zero but less than 0.3 (well actually any thing less than 1/3 will do). Then show that for all delta>0 there is always some x such that |x-1|<delta and |x/(x+1) - 1|>epsilon.

CB

Re: prove limit by definition

what is the problem of my way.

??

Re: prove limit by definition

Consider $\displaystyle {\delta}$ <_1. Then if x satisfy's 0< |x-2|< $\displaystyle {\delta}$, F(x)<3/4. Hence |F(x)-1|>1/4. Choosing $\displaystyle {\epsilon}$<_1/4 completes the proof.

Re: prove limit by definition

i cant because its for every delta

and can you tell me where am i wrong

?

Re: prove limit by definition

For any delta>1, x values in (1,3) will still have to satisfy |F(x)-1|< epsilon.

Re: prove limit by definition

It's not true that you are totally free to choose any ε that you like.

As **CaptainBlack** told you, the actual limit is 2/3.

|1 - 2/3| = 1/3. So, if you use an ε ≥ 1/3, it **is** possible to find a δ > 0 such that for all x for which 0 < | x - 2 | < δ it follows that | x/(x+1) - 1 | < ε .

That's why **CB** said you need to choose ε < 1/3 .