text: # Thread: show that f(x) has a removable discontinuity...
  
  1. ## 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)
  
  $$f(x)=\left\{\begin{array}{\frac{3}{x}+\frac{-2x + 6}{x(x-2)}}, &\mbox{ if }\x \ne 0, 2\\9, & \mbox{if}x=0\end{array}\right[tex] 2. ## Re: show that f(x) has a removable discontinuity... Originally Posted by dokrbb I have the function f(x) equal to 3/x+(-2x + 6)/[x(x-2)] when x different to 0, and 2, and f(x) equals 9 when x = 0 I have to show that f(x) has a removable discontinuity at x=0 I'm stucked with the transformation of these fractions and it leads me nowhere, can you show me please the first few steps? I also tried to put it in LaTeX but I got something wrong, (see below) \displaystyle f(x)=\left\{\begin{array}{\frac{3}{x}+\frac{-2x + 6}{x(x-2)}}, &\mbox{ if }\x \ne 0, 2\\9, & \mbox{if}x=0\end{array}\right If the limit from the left of zero is equal to the limit from the right of zero AND that limit is a finite number then the function will have a removeable discontinuity. Calculate the 1 sided limits as x--> 0 3. ## Re: show that f(x) has a removable discontinuity... Try [tex]f(x)=\begin{cases}{\frac{3}{x}+\frac{-2x + 6}{x(x-2)}}, &\mbox{ if }x \ne 0, 2\\9, & \mbox{ if }x=0\end{cases}\right$$
  
  It comes out as
  $\displaystyle f(x)=\begin{cases}{\frac{3}{x}+\frac{-2x + 6}{x(x-2)}}, &\mbox{ if }x \ne 0, 2\\9, & \mbox{ if }x=0\end{cases}\right$
  
  I don't recall how to center the 9.
  
  -Dan
  
  4. ## Re: show that f(x) has a removable discontinuity...
  
  Thanks to both of you - I've got the answer, It's $\displaystyle \frac{1}{-2}$