Question

Find the following limits:
a)f(x)= {6-x^2 if x1 }, lim(x->1) f(x)=

Answer 1

piecewise function right?
this is much easier than you think
replace \(x\) by 1 in both expressions
if you get the same answer, that is the limit
if you do not get the same answer, there is no limit

Answer 2

\[f(x)= \left\{\begin{array}{rcc}
6-x^2& \text{if} & x\leq 1 \\
2x-1& \text{if} & x >1
\end{array}
\right.
\]
in the first one you get \(6-1=5\) in the second you get \(2-1=1\) and since \(5\neq 1\) there is no limit

Answer 3

oh ok!

Answer 4

I have more can you help?

Answer 5

sure ask away !

Answer 6

btw the last one was easy when you knew what to do, yes?

Answer 7

yeah i know i just get confused on how to go about doing them

Answer 8

\[\lim_{x \rightarrow 2} (\sqrt{2x+2}-\sqrt{6}) \div 2x-4\]

Answer 9

it is that part in parentheses over the part behind the division sign but i didnt know how to draw that

Answer 10

first try replacing \(x\) by 2
unfortunately if you do this, you get \(\frac{0}{0}\) so we will have to do more work

Answer 11

\[\lim_{x\to 2}\frac{\sqrt{2x+2}-\sqrt{6}}{2x-4}\] is your question yes?

Answer 12

Yeah!
sorry my computer was freaking out

Answer 13

ok here is what you need to do
multiply top and bottom by the "conjugate" of the numerator
the conjugate of \(\sqrt{2x+2}-\sqrt{6}\) is \(\sqrt{2x+2}+\sqrt{6}\)
it is similar to "rationalizing the denominator" except this time you are rationalizing the numerator

Answer 14

uh huh

Answer 15

leave the denominator in parentheses
the reason this works is that \((\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b\)
you get
\[\frac{\sqrt{2x+2}-\sqrt{6}}{2x-4}\times \frac{\sqrt{2x+2}+\sqrt{6}}{\sqrt{2x+2}+\sqrt{6}}\]

Answer 16

\[\frac{(2x+2-6}{(2x-4)(\sqrt{2x+2}+\sqrt{6})}=\frac{2x-4}{(2x-4)(\sqrt{2x+2}+\sqrt{6})}\]
cancel, then replace \(x\) by 2

Answer 17

let me know what you get
hope the steps are clear

Answer 18

1/0

Answer 19

no

Answer 20

1/sqrt 12

Answer 21

careful here !!

Answer 22

\[\sqrt{6}+\sqrt{6}\neq \sqrt{12}\]

Answer 23

?

Answer 24

don't do all that work and then mess up with simple combining like terms, or your teacher will think you are not ready for the class.
\[\sqrt{6}+\sqrt{6}=2\sqrt{6}\]
so your answer is \(\frac{1}{2\sqrt{6}}\)

Answer 25

clear or no?

Answer 26

oh!

Answer 27

whew
like
\[\sqrt{25}+\sqrt{25}=5+5=10\] not
\[\sqrt{25}+\sqrt{25}=\sqrt{50}\]!!!

Answer 28

yeah i get it now

Answer 29

good!

Answer 30

is that the limit or do we have to do something to get rid of the sqrt in the denominator

Answer 31

no that is the limit
the trick was to get it out of looking like \(\frac{0}{0}\) by cancelling

Answer 32

that is always the trick

Answer 33

oh
ok well the next one is :
\[\lim_{x \rightarrow -2} (x ^{2}-5x+3)/ x-1=\]

Answer 34

working on it

Answer 35

what happens when you replace \(x\) by \(-2\) ?

Answer 36

whatever you get, that is your answer

Answer 37

because the denominator will be \(-2-1=-3\) and so long as the denominator is not zero, you can just evaluate

Answer 38

17/-3

Answer 39

correct?

Answer 40

i don't know i didn't do it
but the -3 is right
give me a second

Answer 41

yeah numerator is 17

Answer 42

yay! ok i started the next one and it comes out to 0/0 and im not sure where to go from there:
\[\lim_{x \rightarrow 2} ( x ^{2}-5x+4) / x-4\]

Answer 43

ok gimmick is as before
factor and cancel

Answer 44

is the answer 3?

Answer 45

and don't think too hard in factoring
since if you replace \(x\) by 4 in the numerator, you know it must factor as
\[x^2-5x+4=(x-4)(\text{something})\]

Answer 46

factor, cancel, you get \(x-1\) replace \(x\) by 4 and get 3
yes, answer is 3

Answer 47

yay! im actually getting it this is a big thing for me! i am terrible at math but i have to take this so i am trying my best to get through it

Answer 48

\[\lim_{x \rightarrow -4} (3x+12)/\left| x+4 \right|\]

Answer 49

forget this one

Answer 50

ok i got 0/0 to begin with but with the absolute values does that mean that the -4 turns positive?

Answer 51

\(|x+4|\) is a piecewise function, so the limit from the left and from the right will be different

Answer 52

no it does not

Answer 53

did you mean forget about this one? like as in the absolute value?

Answer 54

i mean forget about it having a limit

Answer 55

it will not exist
here is a nice picture to show you why
http://www.wolframalpha.com/input/?i=%283x%2B12%29%2F |x%2B4|
notice the nice vertical line at \(x=-4\)

Answer 56

you will have to cut and paste not click
absolute value signs ruin it for some reason

Answer 57

oh ok

Answer 58

ok well i still have more if you are not tired of helping me yet :)

Answer 59

the reason is this
\(|x+4|\) is a piecewise function
if \(x<-4\) then \(|x-4|=-x-4\) and if \(x>-4\) \(|x+4|=x+4\)
you are trying to take the limit as \(x\to -4\) but the function changes definition there

Answer 60

go ahead and post, but put it in a new thread, this is getting too long