anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@amriju
anonymous
  • anonymous
@uri @John_ES
anonymous
  • anonymous
you sure you have the problem stated correctly?

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More answers

anonymous
  • anonymous
Yes, I can reword it because it isn't stated the way it is in the problem
anonymous
  • anonymous
\[f(x)=\left\{ \frac{ 1+|3-x| if x \neq3 }{ 2 if x =3 } \right\}\]
anonymous
  • anonymous
DNE 3+ = 2 3 - = DNE 2 not equal to DNE, so the limit doesnot exists
anonymous
  • anonymous
For all 3 equations things then?
anonymous
  • anonymous
for X--->3= DNE X can never be equal to 3 so 3^- DNE when X--->3
anonymous
  • anonymous
no... limit as x ->3^+ = limit as x ->3^- = 2
anonymous
  • anonymous
Alright well how about this problem then \[\lim_{x \rightarrow 3^-} x \sqrt{9-x^2}\]
anonymous
  • anonymous
0
anonymous
  • anonymous
Yes its zero but how would I put the answer?
anonymous
  • anonymous
SO the complete equations for the first problem I asked would become DNE because of the 2 with x=3
anonymous
  • anonymous
yes
anonymous
  • anonymous
expand it for example sqrt (3-x)^2
anonymous
  • anonymous
no...
anonymous
  • 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
  • anonymous
(3-x)= 3-3 = 0
anonymous
  • 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
  • anonymous
the first part doesnt exist since X can not be 3
anonymous
  • anonymous
f(3) = 2 but limit f(x) as x -> 3 = 1.
anonymous
  • anonymous
You guys are confusing me.
anonymous
  • anonymous
they are not equal to eachothe therefore, X-> 3= DNE
anonymous
  • anonymous
Thank you! And then for the second equation we wouldn't put the lim is 0. Wouldn't that be the indeterminate
anonymous
  • anonymous
is this your function?\[f(x) =\left[\begin{matrix}1+|3-x| & x\neq 3 \\ 2& x=3\end{matrix}\right] \]
anonymous
  • anonymous
Yes it is
anonymous
  • anonymous
and you want this?\[\lim_{x \rightarrow 3} f(x)\]
anonymous
  • anonymous
No I want. |dw:1378745223628:dw|
anonymous
  • anonymous
a) 1 b) 1 c) 1
anonymous
  • anonymous
Your saying that the only lim factor they have is a 1? Not DNE?
anonymous
  • anonymous
Does @helpme1.2 think that as well?
anonymous
  • 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
  • anonymous
|dw:1378745407137:dw| this is what your function looks like
anonymous
  • 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
  • anonymous
So how does that function represent 1?
anonymous
  • 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
  • anonymous
Its 3
anonymous
  • anonymous
what? y = 3 as x -> 3?
anonymous
  • anonymous
I dont know
anonymous
  • anonymous
look at the graph!
anonymous
  • anonymous
2
anonymous
  • anonymous
if x isn't 3, what part of the graph are you on?
John_ES
  • 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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