# Problem with precise definition of the limit

• Feb 21st 2010, 09:25 PM
bobey
Problem with precise definition of the limit
Question :

let $\displaystyle h(x)=x^2, x<2$
$\displaystyle = 3, x=2$
$\displaystyle =2, x>2$

Show that
lim(x-->2)h(x)does not equal to 4

my teacher give the answer :

for $\displaystyle 2<x<2+delta => h(x) = 2 => abs(h(x)-4)=2$

Thus for $\displaystyle epsilon < 2, abs(h(x)-4) >= epsilon$ whenever $\displaystyle 2<x<2+delta$ no matter how small we choose
$\displaystyle delta > 0$ => lim(x-->2)h(x)does not equal to 4

i cannot understand the second line of the explanation. why we have to consider $\displaystyle epsilon < 2$

why cannot consider this : $\displaystyle abs(h(x)-4)=2 < epsilon$ thus, $\displaystyle epsilon should > 2$ right?

can anyone explain it to me? i really could'nt understand this!(Shake)
• Feb 22nd 2010, 12:48 AM
HallsofIvy
In order that the limit be 4, you must be able to make $\displaystyle |f(x)- 4|< \epsilon$ for any $\displaystyle \epsilon> 0$. Your teacher was pointing out that the difficulty (the fact that you can't make it "$\displaystyle <\epsilon$" happens if you take $\displaystyle \epsilon< 2$.

The fact that it is true for some $\displaystyle \epsilon$ ($\displaystyle \epsilon> 2$) doesn't matter. It has to be true for all $\displaystyle \epsilon$.
• Feb 22nd 2010, 01:04 AM
bobey
Quote:

Originally Posted by HallsofIvy
In order that the limit be 4, you must be able to make $\displaystyle |f(x)- 4|< \epsilon$ for any $\displaystyle \epsilon> 0$. Your teacher was pointing out that the difficulty (the fact that you can't make it "$\displaystyle <\epsilon$" happens if you take $\displaystyle \epsilon< 2$.

The fact that it is true for some $\displaystyle \epsilon$ ($\displaystyle \epsilon> 2$) doesn't matter. It has to be true for all $\displaystyle \epsilon$.

can you give some example, i.e. replace the epsilon and delta with a number so that i can 'see' my teacher's explanation. still confuse...

what is the relationship between a delta and epsilon in this case? i can't see the relationship... wat the delta have to do with the epsilon value in this problem???
anyone???
• Feb 23rd 2010, 01:14 AM
bobey
In order to show that lim x->x_0 f(x) does not equal to L, you have to show that

for every $\displaystyle \delta > 0$, there exist $\displaystyle \epsilon > 0$

thus $\displaystyle |x - x_0| < \delta$ and $\displaystyle |f(x) - L|>\epsilon$

so in the above case

$\displaystyle |x-2|<\delta$ ==> $\displaystyle 2-\delta<x<2+\delta$

and thus $\displaystyle 2<x<2+\delta$ which is from the graph

==> h(x) = 2 ==> $\displaystyle |h(x) - L|$ = $\displaystyle |2-4|$ = $\displaystyle |-2|$ = $\displaystyle 2 >\epsilon$

if we take $\displaystyle \epsilon = 0.0001$,thus $\displaystyle |h(x) - L|> \epsilon$ (since $\displaystyle \epsilon > 0$

thus, it shows that lim(x-->2)h(x)does not equal to 4