Question

Are these correct?

Answer 1

@Nnesha
@ganeshie8

Answer 2

4 is correct.

Answer 3

How is that one correct? I got the same answers as the first 3 problems but not number 4.

Answer 4

I believe your goal is to plug in x, the x is stated for you.
(2)^2 + 5(2) + 6
--------------
(2)+2

4+10+6
---------
4


20/4 = 5. Doing this in my head, apologies for any error.

Answer 5

Oh I made a mistake in that. Thanks! Can you explain number 5 too @Mikeyy1992?

Answer 6

I shall try, one sec.

Answer 7

Okay

Answer 8

x-1
------
x^2 - 1

X is stated to be 1.
1-1
------
1^2 - 1

0
-

1*1 - 1



0
-
0


The answer shall be 0, if I am not mistaken.

Answer 9

Okay I got that too but my teacher put 1/2.

Answer 10

Do you know how to solve 6 and 7? Sorry if I'm asking for too much.

Answer 11

@Mikeyy1992

Answer 12

I can do them in my head, but I do not know how to fully explain 6 and 7, but from my knowledge, the answers are both correct.
For 5, I have dealt with the same exact puzzle, so I know the answer is 0, anyways.

Answer 13

Can you write the steps on how you solve it? I'll see if I can understand when I look at the steps.

Answer 14

6, already proves it will be DNE, with the second statement of x>2.
With number 7, I see cos, so already think negative wise, from past learnings, which I know by heart. It's more so, making e to -e then x/3 into a factor which produces -e^3. I am not good at explaining some terms in mathematics, apologies.

Answer 15

Why does the second statement x>2 prove that it's DNE?

Answer 16

Zepdrix, might point you in a better direction than me; if I am making any mistakes.

Answer 17

Oooo it's Princess Jasmin!! :D

Answer 18

Are these answers that the teacher posted?
Or are we checking YOUR answers?

Answer 19

No it's my teacher's answers and haha ya that's me! :)

Answer 20

For number 6,
remember that for a limit to exist `at a particular point`,
The limits from each direction must match
\(\large\rm x\to2^{+}\) needs to give the same as \(\large\rm x\to2^{-}\)

Answer 21

Gosh, thank you for stating that way better than me, zepdrix! Knew you could! (:

Answer 22

\[\large\rm \lim_{x\to2^{-}}2x^2-4x\]So if you look at this big ole piece-wise function...
when we approach 2 from the `left side`, we're using the top piece, understand why? :o

Answer 23

The top piece meaning 2x^2 -4x?

Answer 24

ya

Answer 25

We are trying to see if the function is indeterminate or if the point doesn't exist? Or if the function is continuous or if it has a hole?

Answer 26

For piece-wise functions,
we need to make sure that the `limit from the left` agrees with the `limit from the right`.
If they don't agree, then the limit does not exist at that point.

So we have some big piece-wise function, let's call it f(x).
We want to know whether or not \(\large\rm \lim_{x\to2}f(x)\) exists, and if so, find the value.
To determine this, we need to check and see if the left and right limits agree,\[\large\rm \lim_{x\to2^-}f(x)\stackrel{?}{=}\lim_{x\to2^+}f(x)\]

Answer 27

So again I'll ask,\[\large\rm \lim_{x\to2^{-}}f(x)=\lim_{x\to2^{-}}2x^2-4x\]Do you understand why I'm replacing f(x) with this top line in the piece-wise function for x approaching from the left?

Answer 28

Okay so wouldn't you just insert in a 2 on both sides of the equation? It would still give the same answer right?

Answer 29

Yes, if that's easier, maybe just think of it that way :)
The pieces need to give the same output at x=2.

Answer 30

\[\large\rm 2x^2-4x \stackrel{?}{=}4\sin\left(\frac{\pi x}{4}\right)\]So are these equal when you plug in x=2? :)

Answer 31

0 = 3.99999

Answer 32

So no it doesn't match.

Answer 33

Oh so do you that process for all inequalities that pop up with the lim function?

Answer 34

These were our steps:
\[\large\rm \lim_{x\to2^-}f(x)\stackrel{?}{=}\lim_{x\to2^+}f(x)\]\[\large\rm \lim_{x\to2^-}2x^2-4x\stackrel{?}{=}\lim_{x\to2^+}4\sin\left(\frac{\pi x}{4}\right)\]\[\large\rm 2(2)^2-4(2)\stackrel{?}{=}4\sin\left(\frac{4\pi}{4}\right)\]\[\large\rm 0\ne4\]I suppose yes.
It might get a tiny bit more complicated if your teacher throws piece-wise functions at you with 3 or 4 pieces. But it should always be the same process :O

Answer 35

Should we go over number 5? :o
looks like you were both a little confused on that one.

Answer 36

Ya can you go over that and number 7? Sorry!

Answer 37

\[\large\rm \lim_{x \to1}\frac{x-1}{x^2-1}=\frac{0}{0}\ne0\]0/0 is an indeterminate form, it's not equal to 0.
So we have to work our way around it some how.

Answer 38

Notice that the denominator is the difference of squares,\[\large\rm x^2-1=x^2-1^2\]We have a rule for factoring the difference of squares, do you remember it? :)

Answer 39

So we have to multiply both top and bottom by x+1?

Answer 40

Umm sure, that's another option!
That's pretty much the same thing I was doing, but in reverse.

Answer 41

Oh okay. That's how my teacher taught me. So after reducing or crossing out you would get 1/(x+1)?

Answer 42

And then you would have to plug in 1 right?

Answer 43

Yes :)

Answer 44

Oh, that's what I forgot, been a couple years! :P Thank you again, Zepdrix.

Answer 45

Ohhhh that's why the answer is 1/2!

Answer 46

So this is what you do with limits.
Step 1: You try to plug the number directly in.
~If you run into trouble, you have to try and `cancel stuff out`.
Step 2: Cancel stuff out.
Step 3: Try to plug the number directly in again.

Answer 47

Okay I understood that. Thanks! Can you help me with number 7?

Answer 48

See the steps I listed?
Number 7 is going to be a piece of cake for us.
We'll apply step 1, and we won't run into any problems.
It will simplify down to an answer.

Answer 49

Plug 3 directly into the function.
We get something like this,\[\large\rm e^{3}\cos\left(\frac{\pi(3)}{3}\right)\]yes?

Answer 50

And now simplify :)

Answer 51

\[e^3\cos(3.14) = e^3(-1) = -e^3\]

Answer 52

You and your ways -_- lol
You gotta remember your trig stuff silly billy!
\(\large\rm \cos(\pi)=-1\) <- yes. I don't like the 3.14 though lol

Answer 53

Lol sorry! But thank you soo much!! You are really good at helping others! :D

Answer 54

yay team \c:/ we did it!

Answer 55

Thanks @Mikeyy1992! You were very helpful too! :)