Question

lim
x-> 4(-) *coming from the left*

(√x -2)/(x-4) please show steps

Answer 1

whats under the sq. rt

Answer 2

its a the square root of x

Answer 3

factor the bottom

Answer 4

\[\frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\]

Answer 5

so we have
\[\lim_{x \rightarrow 4^-}\frac{1}{\sqrt{x}+2}=\frac{1}{\sqrt{4}+2}=\frac{1}{4}\]

Answer 6

oh so you factored the bottom by √x because √x times itself cancels out to be just x right?

Answer 7

yes

Answer 8

you can also multiply top and bottom by the conjugate of the top

Answer 9

this will also work

Answer 10

can you help me with another one, this one involving piecewise?

Answer 11

satellite will help you

Answer 12

oooh can i write the piecewise function? i mean the latex, not the answer

Answer 13

lim
x->2 where f(x)= { x^2 -4x + 6 X<2
-x^2 +4x-2 x (greater than or equal to) 2

Answer 14

plug in 2 into both expressions
if you get the same output,then the limit is that
if you get different outputs, then the limit is undefined

Answer 15

\[f(x) = \left\{\begin{array}{rcc}
x^2 + 4x+6 & \text{if} & x <2 \\
- x^2+ 4-2& \text{if} & x \geq 2\
\end{array}
\right.\]

Answer 16

so for the first one x is 18, which doesn't fit with the first set therefore undefined? and the second is 2 which fits so the limit would be 2?

Answer 17

tada!

Answer 18

what myininaya said. plug in 2
get different numbers
no limit

Answer 19

oh so the limit would be 2

Answer 20

yes

Answer 21

oh satellite missed an x

Answer 22

i will never use satellite's expressions again

Answer 23

lol

Answer 24

and a negative sign

Answer 25

\[f(x) = \left\{\begin{array}{rcc}
x^2 - 4x+6 & \text{if} & x <2 \\
- x^2+ 4x-2& \text{if} & x \geq 2\
\end{array}
\right.\]

Answer 26

\[2^2-4(2)+6=2=2=-(2)^2+4(2)-2\]
so yes the limit is 2

Answer 27

thank you all. can i ask you all something? how many years of math did you take in college if you did attend?

Answer 28

omg several+some more

Answer 29

10

Answer 30

4 undergrad...6 grad

Answer 31

Well I'm trying to advance in math and be as good as you guys

Answer 32

your help is much appreciated

Answer 33

i am so ashamed
\[f(x) = \left\{\begin{array}{rcc}
x^2 -4x+6 & \text{if} & x <2 \\
- x^2+ 4x-2& \text{if} & x \geq 2\
\end{array}
\right.\]

Answer 34

lol

Answer 35

oops the units is years sorry
5 to 6 something liek that

Answer 36

undergrad
0
grad
\[\infty\]

Answer 37

satellite i was using your function at first

Answer 38

unfortunately sometimes
\[0+\infty=0\] unlike what you may have heard

Answer 39

yeah i messed up so i take the blame

Answer 40

must be the 'new math'

Answer 41

maybe it is
\[\frac{0}{\infty}\]

Answer 42

i can't read this before tomorrow and know what i'm talking about
i think i will go to sleep

Answer 43

night

Answer 44

i'm reading something i know nothing about and my adviser wants me lead a discussion on it tomorrow

Answer 45

good luck!

Answer 46

at least its informal

Answer 47

that's good

Answer 48

goodnight guys

Answer 49

one more question im doing this continuity and discontinuity at points and which are the points where it isnt continuous. it says which of the discontinuities are removable? what exactly does that mean by the question?