find limits if they exist f(x)={1+|3-x| if x is unequal to 3}
{2 if x=3}
a.) lim x----->3^+
b.)lim x----->3^-
c.)lim x ----> 3

- anonymous

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- anonymous

@amriju

- anonymous

@uri
@John_ES

- anonymous

you sure you have the problem stated correctly?

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## More answers

- anonymous

Yes, I can reword it because it isn't stated the way it is in the problem

- anonymous

\[f(x)=\left\{ \frac{ 1+|3-x| if x \neq3 }{ 2 if x =3 } \right\}\]

- anonymous

DNE
3+ = 2
3 - = DNE
2 not equal to DNE, so the limit doesnot exists

- anonymous

For all 3 equations things then?

- anonymous

for X--->3= DNE
X can never be equal to 3 so 3^- DNE when X--->3

- anonymous

no...
limit as x ->3^+ = limit as x ->3^- = 2

- anonymous

Alright well how about this problem then
\[\lim_{x \rightarrow 3^-} x \sqrt{9-x^2}\]

- anonymous

0

- anonymous

Yes its zero but how would I put the answer?

- anonymous

SO the complete equations for the first problem I asked would become DNE because of the 2 with x=3

- anonymous

yes

- anonymous

expand it for example sqrt (3-x)^2

- anonymous

no...

- anonymous

my computer is having issues. if your function is 1 + |3 - x| when x isn't 3 and 2 when x is 3 then the limit as x goes to 3 is 2.

- anonymous

(3-x)= 3-3 = 0

- anonymous

oops...
the limit of the first part is 1 and the second part is 2. it's not continuous but the limit does exist and the limit is 1.

- anonymous

the first part doesnt exist since X can not be 3

- anonymous

f(3) = 2 but limit f(x) as x -> 3 = 1.

- anonymous

You guys are confusing me.

- anonymous

they are not equal to eachothe therefore, X-> 3= DNE

- anonymous

Thank you! And then for the second equation we wouldn't put the lim is 0. Wouldn't that be the indeterminate

- anonymous

is this your function?\[f(x) =\left[\begin{matrix}1+|3-x| & x\neq 3 \\ 2& x=3\end{matrix}\right] \]

- anonymous

Yes it is

- anonymous

and you want this?\[\lim_{x \rightarrow 3} f(x)\]

- anonymous

No I want. |dw:1378745223628:dw|

- anonymous

a) 1
b) 1
c) 1

- anonymous

Your saying that the only lim factor they have is a 1? Not DNE?

- anonymous

Does @helpme1.2 think that as well?

- anonymous

yes... the function is not continuous because limit f(x) as x -> 3 does not equal f(3) but the limit f(x) as x ->3 is 1

- anonymous

|dw:1378745407137:dw|
this is what your function looks like

- anonymous

From the left or right side? Do you even know what the + or - means behind the 3? the + means that it goes to the right side, if it is the - it goes to the left side.

- anonymous

So how does that function represent 1?

- anonymous

what does the function go to when x approaches 3? that means what is y when x approaches 3. remember, it doesn't get to 3 but can get as close as you want. what is y? from the left and the right. look at the graph.

- anonymous

Its 3

- anonymous

what? y = 3 as x -> 3?

- anonymous

I dont know

- anonymous

look at the graph!

- anonymous

2

- anonymous

if x isn't 3, what part of the graph are you on?

- John_ES

There is another way to obtain the solution, although without the usefull graphics. We can rewrite the expression of the function without absolute values in this way,
\[g(x)=1+|3-x|\Leftrightarrow g(x)=\begin{cases} 1+(3-x)\ \ \text{if} \ \ x<3\\ 1-(3-x)\ \ \text{if}\ \ x>3\end{cases}\]So,
\[g(x)=\begin{cases} 4-x\ \ \text{if} \ \ x<3\\ -2+x\ \ \text{if}\ \ x>3\end{cases}\]Put in the original function, we have,
\[f(x)=\begin{cases}4-x\ \ \text{if}\ \ x<3\\ -2+x\ \ \text{if}\ \ x>3\\ 2\ \ \text{if}\ \ x=3\end{cases}\]
The first line, let us evaluate limits for 3^-, the second line limits for 3^+. So,
\[\lim_{x\rightarrow3^-}f(x)=\lim_{x\rightarrow3}(4-x)=1\]\[\lim_{x\rightarrow3^+}f(x)=\lim_{x\rightarrow3}(-2+x)=1\]
The last limit only exists if the lateral limits exist and are equal, as in this case happens. Then its value is just the value of the lateral limits.
\[\lim_{x\rightarrow3}f(x)=1\]But you can check that this function is not continuous because f(3)=2, and doesn't take the same value of the lateral limits.

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