Question

HOW?!
Find the indicated limit, if it exists.

Answer 1

@SolomonZelman pleeease...

Answer 2

ok, sure

Answer 3

In general:

\(\Large\color{slate}{\displaystyle\lim_{x \rightarrow ~a}f(x)}\) exist, if and only if \(\Large\color{slate}{\displaystyle\lim_{x \rightarrow ~a^+}f(x)=\lim_{x \rightarrow ~a^-}f(x)}\)

Answer 4

(( this is for any value of a, that x (or another variable) is approaching in any function ))

Answer 5

okay..

Answer 6

Same way, if \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0^{-}}f(x)\ne\lim_{x \rightarrow ~0^{+}}f(x)}\) then the limit would not exist.

Answer 7

I am saying that [a two-sided] limit \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}f(x)}\) wouldn't exist.

Answer 8

answer 2 questions for me

Answer 9

i'll try lol

Answer 10

Based on the picture, what is \(\Large\color{slate}{\displaystyle\lim_{x \rightarrow ~0^{\color{blue}{+}}}f(x)}\) ?

Based on the picture, what is \(\Large\color{slate}{\displaystyle\lim_{x \rightarrow ~0^{\color{blue}{\LARGE -}}}f(x)}\) ?

Answer 11

I mean based on the function

Answer 12

the limit from the right side is when x>0
the limit from the left side is when x<0

Answer 13

okay first picture or second picture??

Answer 14

this one you posted

Answer 15

\(\LARGE\color{slate}{\displaystyle\lim_{x \rightarrow ~0^{\color{blue}{+}}}f(x)=\color{blue}{-4(0)+9}=9}\)

\(\LARGE\color{slate}{\displaystyle\lim_{x \rightarrow ~0^{\color{red}{-}}}f(x)=\color{red}{9-(0)^2}=9}\)

Answer 16

the part of the function where x>0, I use to find the right-side limit, and the part of the function where x<0 I use to find the left-side limit.

Answer 17

i wanna say 9 but i think i'm wrong..

Answer 18

9 as the final answer?

Answer 19

then, it is correc

Answer 20

ct

Answer 21

really?? :D

Answer 22

what about the other one?? -4??

Answer 23

I did the second problem.... you noticed, right?

Answer 24

yes, sorry on my test they are reverses lol

Answer 25

reversed*

Answer 26

anyway, we are done with the problem where x approaches 0?

Answer 27

well i mean i think so..

Answer 28

ok, will move on to next one

Answer 29

\(\LARGE\color{black}{ f(x) = \begin{cases} & x+3,~~~{\large {\rm if~~}x<-4} \\ & 3-x,~~~{\large {\rm if~~}x\ge -4} \end{cases} }\)

Answer 30

okay :)

Answer 31

I will indicate in colors what limit corresponds to what part of the function.

\(\LARGE\color{black}{ f(x) = \begin{cases} & \color{red}{x+3},~~~{\large {\rm if~~}x<-4} \\ & \color{blue}{3-x},~~~{\large {\rm if~~}x\ge -4} \end{cases} }\)


\(\LARGE \color{blue}{\displaystyle\lim_{~~~~~~~~~~x \rightarrow ~(-4)^+}f(x)}\)

\(\LARGE \color{red}{\displaystyle\lim_{~~~~~~~~~~x \rightarrow ~(-4)^-}f(x)}\)

Answer 32

alright....

Answer 33

Ok, find each of the limits

\(\LARGE \color{blue}{\displaystyle\lim_{~~~~~~~~~~x \rightarrow ~(-4)^+}f(x)}\) and \(\LARGE \color{red}{\displaystyle\lim_{~~~~~~~~~~x \rightarrow ~(-4)^-}f(x)}\)

Answer 34

...-4..??

Answer 35

I don't think any of the limits is equal (approaching a y-value of) -4.

Answer 36

okay so in this case the limit does not exist...???

Answer 37

I am not saying right or wrong, but why do you think this way?

Answer 38

(don't be afraid, and don't hesitate)

Answer 39

idk... i just makes since because if the limit isn't approaching anything... well yeah...

Answer 40

you kinda hit the point, but needed more power.

here is the explanation

Answer 41

\(\LARGE\color{black}{ f(x) = \begin{cases} & \color{red}{x+3},~~~{\large {\rm if~~}x<-4} \\ & \color{blue}{3-x},~~~{\large {\rm if~~}x\ge -4} \end{cases} }\)


\(\LARGE \color{blue}{\displaystyle\lim_{~~~~~~~~~~x \rightarrow ~(-4)^+}f(x)~=~3-(-4)=7}\)

\(\LARGE \color{red}{\displaystyle\lim_{~~~~~~~~~~x \rightarrow ~(-4)^-}f(x)=(-4)+3=-1}\)

So, the graph approaches the y-value of -1, as x approaches -4 from the left (when x is little less than -4, like x=-5, x=-4.5, x=-4.1, x=-4.032 and etc...)

So, the graph approaches the y-value of 7, as x approaches -4 from the right(when x is little more than -4, like x=-3, x=-3.5, x=-3.7, x=-4.023 and etc...)

{{ NOTE: the numbers I choose in parenthesis, are just to make a point. }}

therefore, a two-sided limit doesn't exist, because the right and left side aren't equivalent.

Answer 42

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~(-4)^+}f(x)\ne\lim_{x \rightarrow ~(-4)^-}f(x)}\) therefore, \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~(-4)}f(x)~~~{\bf DNE}}\)

Answer 43

okay that makes since. THANK YOU SOOOO MUCH!!!!!!!!!!!!!

Answer 44

yeah, sorta tryin.... tnx

Answer 45

if you have some questions about this prob, then ask....

Answer 46

but if not, then have a good evening....

Answer 47

I'm doing a practice exam in few minutes and the real one soon so i'll probably need more help. X) lol thanks!!!!!!!

Answer 48

if I be online, I will do my best....

Answer 49

wait!

Answer 50

I am waitin