continious function

**dopi** · November 14th 2006, 11:24 AM

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

**topsquark** · November 14th 2006, 11:46 AM

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

**dopi** · November 14th 2006, 11:57 AM

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?

**Plato** · November 14th 2006, 12:06 PM

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

Now it is continuous.

**dopi** · November 14th 2006, 12:21 PM

Originally Posted by Plato

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

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

**ThePerfectHacker** · November 14th 2006, 01:04 PM

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

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

**dopi** · November 14th 2006, 01:08 PM

Originally Posted by ThePerfectHacker

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

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

**topsquark** · November 14th 2006, 05:30 PM

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 $ln | \pm 1 | = 0$ so $\frac{1}{ln|x|}$ is undefined at $x = \pm 1$ .

-Dan

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