# Thread: continious function

1. ## continious function

f:[-1,1] -> R is defined as

.......| 1/ln(x) , x not equal be 0
f(x)={
.......| 0, x=0

how do i show this is continious???
thankz

2. Originally Posted by dopi
f:[-1,1] -> R is defined as

.......| 1/ln(x) , x not equal be 0
f(x)={
.......| 0, x=0

how do i show this is continious???
thankz
But ln(x) is only defined on (0, infinity)! Your function's domain is too large.

-Dan

3. ## ??

Originally Posted by topsquark
But ln(x) is only defined on (0, infinity)! Your function's domain is too large.

-Dan
no its not ...its defined on [-1, 1] isnt it?

4. Maybe you mean:
$\displaystyle f(x) = \left\{ {\begin{array}{lc} {\frac{1}{{\ln \left( {\left| x \right|} \right)}}} & {x \not= 0} \\ 0 & x = 0 \\ \end{array}} \right.$

Now it is continuous.

5. Originally Posted by Plato
Maybe you mean:
$\displaystyle f(x) = \left\{ {\begin{array}{lc} {\frac{1}{{\ln \left( {\left| x \right|} \right)}}} & {x \not= 0} \\ 0 & x = 0 \\ \end{array}} \right.$

Now it is continuous.
yea thats wat i meant...i forgot to put the modulus sign on...but how would i show that this is continious with f:[-1,1]
thankz

6. Still no good you should have mentioned the open interval $\displaystyle (-1,1)$

Anyway, the way you show continuity is to show that,
$\displaystyle \lim_{x\to 0}f(x)=f(0)=0$
Since,
$\displaystyle f(x)=\frac{1}{\ln|x|}$ open some open interval except possibly at $\displaystyle x=0$ we have,
$\displaystyle \lim_{x\to 0}\frac{1}{\ln|x|}=0$
Which is true if we consider the left-right limits,
$\displaystyle \lim_{x\to 0^-}\frac{1}{\ln (-x)}=\lim_{x\to 0^+}\frac{1}{\ln x}$
And by using L'Hopital's Rule.

7. Originally Posted by ThePerfectHacker
Still no good you should have mentioned the open interval $\displaystyle (-1,1)$

Anyway, the way you show continuity is to show that,
$\displaystyle \lim_{x\to 0}f(x)=f(0)=0$
Since,
$\displaystyle f(x)=\frac{1}{\ln|x|}$ open some open interval except possibly at $\displaystyle x=0$ we have,
$\displaystyle \lim_{x\to 0}\frac{1}{\ln|x|}=0$
Which is true if we consider the left-right limits,
$\displaystyle \lim_{x\to 0^-}\frac{1}{\ln (-x)}=\lim_{x\to 0^+}\frac{1}{\ln x}$
And by using L'Hopital's Rule.
but its not open, its bounded [-1, 1] thats why i put square brackets...thankz for the help

8. Originally Posted by dopi
but its not open, its bounded [-1, 1] thats why i put square brackets...thankz for the help
TPH was referring to the fact that $\displaystyle ln | \pm 1 | = 0$ so $\displaystyle \frac{1}{ln|x|}$ is undefined at $\displaystyle x = \pm 1$.

-Dan