Question

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

Answer 1

sure

Answer 2

they are usually not nearly as hard as they look

Answer 3

do we really have to sketch them?

Answer 4

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

Answer 5

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.

Answer 6

ok lets start with the first one

Answer 7

Thank you!

Answer 8

the definition of the function changes at 2 right?

Answer 9

Yes

Answer 10

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 \(8-2\times 2\)

Answer 11

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

Answer 12

guess what we do next?

Answer 13

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

Answer 14

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

Answer 15

and yah just plug it in

Answer 16

we also need to check if it goes to 4?

Answer 17

right

Answer 18

plug in 4 as well?

Answer 19

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

Answer 20

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

Answer 21

Okay, thank you for explaining.

Answer 22

Since I have to plug 4 into the functions

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

i am not sure what you are asking

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

give me a second and i will show you a graph

Answer 33

taking me a second, but soon

Answer 34

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

Answer 35

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

Answer 36

yes
picture almost done
damn syntax is killing me

Answer 37

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

Answer 38

oh just click on it

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

\[\color\magenta\heartsuit\]

Answer 45

@misty1212 very good explanations!