show that f(x) has a removable discontinuity...

I have the function f(x) equal to 3/x+(-2x + 6)/[x(x-2)] when x different to 0, and 2,

and f(x) equals 9 when x = 0

I have to show that f(x) has a removable discontinuity at x=0

I'm stucked with the transformation of these fractions and it leads me nowhere, can you show me please the first few steps?

I also tried to put it in LaTeX but I got something wrong, (see below)

[tex]f(x)=\left\{\begin{array}{\frac{3}{x}+\frac{-2x + 6}{x(x-2)}}, &\mbox{ if }\x \ne 0, 2\\9, & \mbox{if}x=0\end{array}\right[tex]

Re: show that f(x) has a removable discontinuity...

Quote:

Originally Posted by

**dokrbb** I have the function f(x) equal to 3/x+(-2x + 6)/[x(x-2)] when x different to 0, and 2,

and f(x) equals 9 when x = 0

I have to show that f(x) has a removable discontinuity at x=0

I'm stucked with the transformation of these fractions and it leads me nowhere, can you show me please the first few steps?

I also tried to put it in LaTeX but I got something wrong, (see below)

If the limit from the left of zero is equal to the limit from the right of zero AND that limit is a finite number then the function will have a removeable discontinuity. Calculate the 1 sided limits as x--> 0

:)

Re: show that f(x) has a removable discontinuity...

Try

[tex]f(x)=\begin{cases}{\frac{3}{x}+\frac{-2x + 6}{x(x-2)}}, &\mbox{ if }x \ne 0, 2\\9, & \mbox{ if }x=0\end{cases}\right[/tex]

It comes out as

I don't recall how to center the 9.

-Dan

Re: show that f(x) has a removable discontinuity...

Thanks to both of you - I've got the answer, It's