Question

A) if f(2) =4 can you conclude anything about the limit of f(x) as x approaches 2? explain your reasoning


B) if the limit of f(x) as x approaches 2 is 4, can you conclude anything about f(2)? explain your reasoning

Answer 1

What is your gut feel? lol

Answer 2

idk

Answer 3

For part (A)
If \[\Large f(2) = 4\]
Then does that mean
\[\Large \lim_{x\rightarrow 2}f(x) = 4 \qquad \color{red}?\]

Answer 4

not necessarily

Answer 5

That's right :)
Can you think of a counterexample?
(HINT: Think piecewise...)

Answer 6

nope....I'm having issues with piecewise functions as well

Answer 7

Well, a most basic example would be
\[\Large f(x) =\left\{\begin{matrix}4&&&x\le 2\\\\0&&&x>2\end{matrix}\right.\]

Answer 8

Now clearly, f(2) = 4, but the limit doesn't exist as x approaches 2 because they approach different values from the left and from the right.

Answer 9

can that be found if the question were graphed

Answer 10

Possibly... IS the question graphed?

Answer 11

no...but i learn better with visuals.... I'm a little slow

Answer 12

Okay, what this piecewise function looks like on a graph is this:
Say this is your Cartesian Plane

Answer 13

And it's clearer to see that f(2) = 4
but the function, as x approaches 2 from the LEFT, goes to 4, but when it approaches from the RIGHT, it goes to zero.
thus, the limit doesn't exist :)

Answer 14

ohhh i get it

Answer 15

Okay?
So part (a) is done? :)

Answer 16

yea....you can say that

Answer 17

I can say it, but we can't really move on until you can say it yourself :P
Are you convinced about our conclusion with part (A) ?

Answer 18

yea i can see it now

Answer 19

Okay, part (B) then... the converse.
Suppose we know that
\[\Large \lim_{x\rightarrow 2}f(x) = 4\]
does it mean that
\[\Large f(2) = 4 \qquad \color{red}?\]

Answer 20

I don't think so

Answer 21

Can you think of a counterexample?

Answer 22

You can think piecewise, but there are answers which I think are slightly prettier :3

Answer 23

like what

Answer 24

Well, first, maybe you want a piecewise function? :P

Answer 25

Because the 'prettier' one is pretty hard to graph :P

Answer 26

Here's the 'prettier' one (the one which isn't piecewise)
\[\Large f(x) = \frac{x^2-4}{x-2}\]

Answer 27

Clearly, f(2) won't exist, since the denominator would be made zero, right?

Answer 28

right

Answer 29

Now, you just have to show that the limit exists and that
\[\Large \lim_{x\rightarrow2}\frac{x^2-4}{x-2}=4\]
Can you do that? :)

Answer 30

HINT: Numerator is factorable...

Answer 31

yes... the top factors to become (x+2) (x-2) so the x-2's cancel and 2+2 equals 4

Answer 32

Yup precisely :)
So for this function
\[\Large f(x)= \frac{x^2 -4}{x-2}\]
The limit as x goes to 2 is indeed 4, but f(2) is not 4 (since it doesn't exist)


That's it ^_^

Answer 33

So actually, the answer to both of your questions is "no conclusion can be made" :)

Answer 34

ohh ok...now i have another question.... how were you able to find the counterexample \[f \left( x \right)= \frac{ x^{2}-4 }{ x-2 }\]

Answer 35

From experience :)

Answer 36

That's like a textbook example showing that \[\Large \lim_{x\rightarrow a}f(x)\]
is not necessarily equal to \[\Large f(a)\]

Answer 37

I have to go now :)
Hope you learned something today XD
------------------------------------------
Terence out

Answer 38

ok thanks..this math lesson has been very helpful