# Thread: Problem with precise definition of the limit

1. ## Problem with precise definition of the limit

Question :

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

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

my teacher give the answer :

for $2 h(x) = 2 => abs(h(x)-4)=2$

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

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

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

can anyone explain it to me? i really could'nt understand this!

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

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

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

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

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???

4. In order to show that lim x->x_0 f(x) does not equal to L, you have to show that

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

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

so in the above case

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

and thus $2 which is from the graph

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

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

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