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 $\displaystyle \frac{1}{-2}$
explanation: The document is a discussion thread from a math forum where a user is asking for help to show that a given function \( f(x) \) has a removable discontinuity at \( x = 0 \). The function is defined piecewise, and the user is having trouble with the algebraic manipulation of the fractions. Another user provides guidance on how to approach the problem by calculating the one-sided limits as \( x \) approaches 0. A third user helps with the LaTeX formatting of the function.
  
  The document contains a question and an answer to the question.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
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distilabel_metadata.role: system
distilabel_metadata.content: # 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 $\displaystyle \frac{1}{-2}$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread from a math forum where a user is asking for help to show that a given function \( f(x) \) has a removable discontinuity at \( x = 0 \). The function is defined piecewise, and the user is having trouble with the algebraic manipulation of the fractions. Another user provides guidance on how to approach the problem by calculating the one-sided limits as \( x \) approaches 0. A third user helps with the LaTeX formatting of the function.
  
  The document contains a question and an answer to the question.
  
  Label: question-and-answer