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

[Joel Jacon]

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?

 

[Jameson]

If I'm understanding correctly this is a piecewise function, correct? Like this?

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

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.?

 

[Joel Jacon]

If I'm understanding correctly this is a piecewise function, correct? Like this?

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

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

 

[Jameson]

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.

 

[Joel Jacon]

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

 

[Jameson]

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:

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