Evaluating $\displaystyle \lim_{{x}\to{2}} f(x$): Does it Exist?

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Discussion Overview

The discussion revolves around evaluating the limit $\displaystyle \lim_{{x}\to{2}} f(x)$ for a piecewise function defined differently for values less than, equal to, and greater than 2. Participants explore the left-hand limit (LHL) and right-hand limit (RHL) to determine if the limit exists.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Post 1 presents the piecewise function and calculates the LHL and RHL, concluding that the limit does not exist.
  • Post 2 confirms the piecewise nature of the function and prompts for specific evaluations of the LHL using values approaching 2 from the left.
  • Post 3 reiterates the piecewise function and seeks clarification on handling the modulus in the context of the LHL.
  • Post 4 discusses the interpretation of modulus as absolute value and suggests evaluating the function at values close to 2.
  • Post 5 continues the discussion on modulus and absolute value, questioning the limits involving $\frac{x}{|x|}$ and their implications for LHL and RHL.
  • Post 6 explains the reasoning behind the limits involving absolute value and relates it back to the original problem of evaluating the limit at 2.

Areas of Agreement / Disagreement

Participants generally agree on the piecewise definition of the function and the need to evaluate the LHL and RHL. However, there is no consensus on the existence of the limit, as different interpretations and calculations are presented.

Contextual Notes

Participants express uncertainty regarding the handling of the modulus (absolute value) in the context of limits, and there are unresolved questions about the implications of these evaluations on the limit's existence.

Joel Jacon
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Evaluate $\displaystyle \lim_{{x}\to{2}} f(x$) if it exist where $f(x)$ = x - |x| where x<2;4 where x = 2;3x - 5 where x>2?

LHL

$\displaystyle \lim_{{x}\to{2}} f(2x)$ = 4

RHL

$\displaystyle \lim_{{x}\to{2}} f(3x - 5)$ = 1

Therefore the limit x tend to 2 for the function does not exist.

Have I done correctly? If not tell me what mistake I made?
 
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If I'm understanding correctly this is a piecewise function, correct? Like this?

$$f(x)=\begin{cases}x-|x|, & x<2 \\[3pt] 4, & x=2 \\[3pt] 3x-5, & x>2 \\ \end{cases}$$

If so, then for the LHL we should look at what function applies when $x<2$. Which one is that? Once you pick that function what value do you get for 1.9, 1.99, 1.999, etc.?
 
Jameson said:
If I'm understanding correctly this is a piecewise function, correct? Like this?

$$f(x)=\begin{cases}x-|x|, & x<2 \\[3pt] 4, & x=2 \\[3pt] 3x-5, & x>2 \\ \end{cases}$$

If so, then for the LHL we should look at what function applies when $x<2$. Which one is that? Once you pick that function what value do you get for 1.9, 1.99, 1.999, etc.?

Can you please tell me how to deal with the modulus. i know that x - |x| is the function to be used for LHL as x<2
 
pcforgeek said:
Can you please tell me how to deal with the modulus. i know that x - |x| is the function to be used for LHL as x<2

I'm not familiar with the term modulus for these brackets, but that might another term for it. I'm interpreting it as absolute value, meaning its distance from 0. You are correct that our $f(x)$ is $x-|x|$. Try plugging in numbers as I suggested. Did you try that?

$(1.9-|1.9|)=(1.9-1.9)=0$

What about 1.99? 1.999?

EDIT: It seems modulus and absolute value are synonymous in this situation. Learn something new every day.
 
Jameson said:
I'm not familiar with the term modulus for these brackets, but that might another term for it. I'm interpreting it as absolute value, meaning its distance from 0. You are correct that our $f(x)$ is $x-|x|$. Try plugging in numbers as I suggested. Did you try that?

$(1.9-|1.9|)=(1.9-1.9)=0$

What about 1.99? 1.999?

EDIT: It seems modulus and absolute value are synonymous in this situation. Learn something new every day.

If modulus is the absolute value then can you explain me why we use $\lim_{{x}\to{0}} \frac{x}{|x|}$ as $\lim_{{x}\to{0}} \frac{x}{-x}$ for LHL
and $\lim_{{x}\to{0}} \frac{x}{x}$ for RHL for the question in which we have to show $\lim_{{x}\to{0}} \frac{x}{|x|}$ does not exist
 
pcforgeek said:
If modulus is the absolute value then can you explain me why we use $\lim_{{x}\to{0}} \frac{x}{|x|}$ as $\lim_{{x}\to{0}} \frac{x}{-x}$ for LHL
and $\lim_{{x}\to{0}} \frac{x}{x}$ for RHL for the question in which we have to show $\lim_{{x}\to{0}} \frac{x}{|x|}$ does not exist

Sure. When we are dealing with positive numbers, or $x>0$, then $|x|=x$. For example |5|=5. When we are dealing with negative numbers, or $x<0$ then |x|=-x. For example |-4|= -(-4)=4.

When are looking at a limit at $x=0$ on the left hand side we have negative numbers and on the right hand side we have positive numbers so when taking the left hand limit we consider negative values of $x$ and when taking the right hand limit we consider positive values of $x$.

Going back to the problem in this thread, what do you get for:

$$\lim_{{x}\to{2^-}}\hspace{1 mm} f(x)$$?
 

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