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

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

Mathematics
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sure
they are usually not nearly as hard as they look

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do we really have to sketch them?
or can we just find the limits if they exist? that is much easier than drawing
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.
ok lets start with the first one
Thank you!
the definition of the function changes at 2 right?
Yes
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 \(8-2\times 2\)
i get \(4\) both times, i bet you do too that means the limit exists, since it is the same in both definitions, and \[\lim_{x\to 2}f(x)=4\]
guess what we do next?
So I can just input all values of two for x? Excluding the third. What exactly makes this limit a piece-wise function?
we are not done yet, we just checked the limit as x goes to 2
and yah just plug it in
we also need to check if it goes to 4?
right
plug in 4 as well?
because only at the change in definition can something go wrong otherwise it is a bunch of polynomials which are always continuous
yeah plug in 4 to the middle one and the bottom one and see if you get the same answer (you do not)
Okay, thank you for explaining.
Since I have to plug 4 into the functions
you are quite welcome told you it was easy right? math teachers often make it seem harder than it is
yeah but only in the middle one and the bottom one where it changes from \[8-2x\] to \[4\]
Since i have to plug 4 into the functions, does that mean that 4 is not part of the limit?
i am not sure what you are asking
if you plug 4 in to \[8-2x\] you get 0 right?
that the limit for the function exists everywhere else but 4. is that the correct?
yes, that's what i got when i plugged it in.
ok and to the right,the function is a constant, it is just 4
so that means, since \(0\neq 4\) you have \[\lim_{x\to 4}f(x)\] does not exist
give me a second and i will show you a graph
taking me a second, but soon
so my final answer would be lim f(x) x->2 and lim f(x) x-4 does not exist
you're fine, take your time. i really appreciate the help.
yes picture almost done damn syntax is killing me
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
oh just click on it
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
and yes, that is your final answer for the limits from the first question graph is what i linked to
okay, thank you so much! that was incredibly helpful! I'll try the next one on my own. :)
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
oh if you want to try it on your own, don't click, although the picture may be helpful
\[\color\magenta\heartsuit\]
@misty1212 very good explanations!

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