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anonymous
 one year ago
Can I please have some help solving this piecewise function/limit?
anonymous
 one year ago
Can I please have some help solving this piecewise function/limit?

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misty1212
 one year ago
Best ResponseYou've already chosen the best response.4they are usually not nearly as hard as they look

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4do we really have to sketch them?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4or can we just find the limits if they exist? that is much easier than drawing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.4ok lets start with the first one

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4the definition of the function changes at 2 right?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4so 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 \(82\times 2\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4i 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
 one year ago
Best ResponseYou've already chosen the best response.4guess what we do next?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I can just input all values of two for x? Excluding the third. What exactly makes this limit a piecewise function?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4we are not done yet, we just checked the limit as x goes to 2

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4and yah just plug it in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we also need to check if it goes to 4?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4because only at the change in definition can something go wrong otherwise it is a bunch of polynomials which are always continuous

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4yeah plug in 4 to the middle one and the bottom one and see if you get the same answer (you do not)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, thank you for explaining.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since I have to plug 4 into the functions

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4you are quite welcome told you it was easy right? math teachers often make it seem harder than it is

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4yeah but only in the middle one and the bottom one where it changes from \[82x\] to \[4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since i have to plug 4 into the functions, does that mean that 4 is not part of the limit?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4i am not sure what you are asking

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4if you plug 4 in to \[82x\] you get 0 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that the limit for the function exists everywhere else but 4. is that the correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, that's what i got when i plugged it in.

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4ok and to the right,the function is a constant, it is just 4

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4so that means, since \(0\neq 4\) you have \[\lim_{x\to 4}f(x)\] does not exist

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4give me a second and i will show you a graph

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4taking me a second, but soon

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so my final answer would be lim f(x) x>2 and lim f(x) x4 does not exist

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you're fine, take your time. i really appreciate the help.

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4yes picture almost done damn syntax is killing me

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4you 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
 one year ago
Best ResponseYou've already chosen the best response.4and yes, that is your final answer for the limits from the first question graph is what i linked to

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, thank you so much! that was incredibly helpful! I'll try the next one on my own. :)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4here 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+%7B1cos%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
 one year ago
Best ResponseYou've already chosen the best response.4oh if you want to try it on your own, don't click, although the picture may be helpful

misty1212
 one year ago
Best ResponseYou've already chosen the best response.4\[\color\magenta\heartsuit\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@misty1212 very good explanations!
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