Can I please have some help solving this piece-wise function/limit?

- anonymous

Can I please have some help solving this piece-wise function/limit?

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- misty1212

sure

- misty1212

they are usually not nearly as hard as they look

- anonymous

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## More answers

- misty1212

do we really have to sketch them?

- misty1212

or can we just find the limits if they exist?
that is much easier than drawing

- anonymous

I'm sure they aren't eventually, I just need more practice with them.
I have to sketch them, if you just show me how to go about finding the limit, I can draw it afterward.

- misty1212

ok lets start with the first one

- anonymous

Thank you!

- misty1212

the definition of the function changes at 2 right?

- anonymous

Yes

- misty1212

so to see if it is continuous at 2, all you need to do is plug in 2 for both definitions of \(f\)
in other words
compute \[2^2\] and \(8-2\times 2\)

- misty1212

i get \(4\) both times, i bet you do too
that means the limit exists, since it is the same in both definitions, and \[\lim_{x\to 2}f(x)=4\]

- misty1212

guess what we do next?

- anonymous

So I can just input all values of two for x? Excluding the third.
What exactly makes this limit a piece-wise function?

- misty1212

we are not done yet, we just checked the limit as x goes to 2

- misty1212

and yah just plug it in

- anonymous

we also need to check if it goes to 4?

- misty1212

right

- anonymous

plug in 4 as well?

- misty1212

because only at the change in definition can something go wrong
otherwise it is a bunch of polynomials which are always continuous

- misty1212

yeah plug in 4 to the middle one and the bottom one and see if you get the same answer (you do not)

- anonymous

Okay, thank you for explaining.

- anonymous

Since I have to plug 4 into the functions

- misty1212

you are quite welcome
told you it was easy right? math teachers often make it seem harder than it is

- misty1212

yeah but only in the middle one and the bottom one where it changes from
\[8-2x\] to \[4\]

- anonymous

Since i have to plug 4 into the functions, does that mean that 4 is not part of the limit?

- misty1212

i am not sure what you are asking

- misty1212

if you plug 4 in to \[8-2x\] you get 0 right?

- anonymous

that the limit for the function exists everywhere else but 4. is that the correct?

- anonymous

yes, that's what i got when i plugged it in.

- misty1212

ok and to the right,the function is a constant, it is just 4

- misty1212

so that means, since \(0\neq 4\) you have
\[\lim_{x\to 4}f(x)\] does not exist

- misty1212

give me a second and i will show you a graph

- misty1212

taking me a second, but soon

- anonymous

so my final answer would be
lim f(x)
x->2
and lim f(x)
x-4 does not exist

- anonymous

you're fine, take your time. i really appreciate the help.

- misty1212

yes
picture almost done
damn syntax is killing me

- misty1212

copy and paste
http://www.wolframalpha.com/input/?i=Plot%5BPiecewise%5B%7B%7Bx%5E2%2C+x+%3C+2%7D%2C+%7B8-2x+%2C+2%3Cx+%3C4%7D%2C%7B4%2Cx%3E4%7D%7D%5D%2C+%7Bx%2C+0%2C+6%7D%5D

- misty1212

oh just click on it

- misty1212

you can see that it is continuous at 2, there is no break there
but there is a jump from 0 to 4 at 4, so not continuous there

- misty1212

and yes, that is your final answer for the limits from the first question
graph is what i linked to

- anonymous

okay, thank you so much! that was incredibly helpful!
I'll try the next one on my own. :)

- misty1212

here is the second one
easier since i had the template
http://www.wolframalpha.com/input/?i=Plot%5BPiecewise%5B%7B%7Bsin%28x%29%2C+x+%3C+0%7D%2C+%7B1-cos%28x%29+%2C+0%3Cx+%3Cpi%7D%2C%7Bcos%28x%29%2Cx%3Epi%7D%7D%5D%2C+%7Bx%2C+-2%2C+6%7D%5D

- misty1212

oh if you want to try it on your own, don't click, although the picture may be helpful

- misty1212

\[\color\magenta\heartsuit\]

- anonymous

@misty1212 very good explanations!

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