Are these correct?
 anonymous
Are these correct?
 Stacey Warren  Expert brainly.com
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 katieb
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 anonymous
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 anonymous
@Nnesha
@ganeshie8
 anonymous
4 is correct.
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 anonymous
How is that one correct? I got the same answers as the first 3 problems but not number 4.
 anonymous
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.
 anonymous
Oh I made a mistake in that. Thanks! Can you explain number 5 too @Mikeyy1992?
 anonymous
I shall try, one sec.
 anonymous
Okay
 anonymous
x1

x^2  1
X is stated to be 1.
11

1^2  1
0

1*1  1
0

0
The answer shall be 0, if I am not mistaken.
 anonymous
Okay I got that too but my teacher put 1/2.
 anonymous
Do you know how to solve 6 and 7? Sorry if I'm asking for too much.
 anonymous
@Mikeyy1992
 anonymous
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.
 anonymous
Can you write the steps on how you solve it? I'll see if I can understand when I look at the steps.
 anonymous
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.
 anonymous
Why does the second statement x>2 prove that it's DNE?
 anonymous
Zepdrix, might point you in a better direction than me; if I am making any mistakes.
 zepdrix
Oooo it's Princess Jasmin!! :D
 zepdrix
Are these answers that the teacher posted?
Or are we checking YOUR answers?
 anonymous
No it's my teacher's answers and haha ya that's me! :)
 zepdrix
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 \(\large\rm x\to2^{}\)
 anonymous
Gosh, thank you for stating that way better than me, zepdrix! Knew you could! (:
 zepdrix
\[\large\rm \lim_{x\to2^{}}2x^24x\]So if you look at this big ole piecewise function...
when we approach 2 from the `left side`, we're using the top piece, understand why? :o
 anonymous
The top piece meaning 2x^2 4x?
 zepdrix
ya
 anonymous
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?
 zepdrix
For piecewise 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 piecewise 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,\[\large\rm \lim_{x\to2^}f(x)\stackrel{?}{=}\lim_{x\to2^+}f(x)\]
 zepdrix
So again I'll ask,\[\large\rm \lim_{x\to2^{}}f(x)=\lim_{x\to2^{}}2x^24x\]Do you understand why I'm replacing f(x) with this top line in the piecewise function for x approaching from the left?
 anonymous
Okay so wouldn't you just insert in a 2 on both sides of the equation? It would still give the same answer right?
 zepdrix
Yes, if that's easier, maybe just think of it that way :)
The pieces need to give the same output at x=2.
 zepdrix
\[\large\rm 2x^24x \stackrel{?}{=}4\sin\left(\frac{\pi x}{4}\right)\]So are these equal when you plug in x=2? :)
 anonymous
0 = 3.99999
 anonymous
So no it doesn't match.
 anonymous
Oh so do you that process for all inequalities that pop up with the lim function?
 zepdrix
These were our steps:
\[\large\rm \lim_{x\to2^}f(x)\stackrel{?}{=}\lim_{x\to2^+}f(x)\]\[\large\rm \lim_{x\to2^}2x^24x\stackrel{?}{=}\lim_{x\to2^+}4\sin\left(\frac{\pi x}{4}\right)\]\[\large\rm 2(2)^24(2)\stackrel{?}{=}4\sin\left(\frac{4\pi}{4}\right)\]\[\large\rm 0\ne4\]I suppose yes.
It might get a tiny bit more complicated if your teacher throws piecewise functions at you with 3 or 4 pieces. But it should always be the same process :O
 zepdrix
Should we go over number 5? :o
looks like you were both a little confused on that one.
 anonymous
Ya can you go over that and number 7? Sorry!
 zepdrix
\[\large\rm \lim_{x \to1}\frac{x1}{x^21}=\frac{0}{0}\ne0\]0/0 is an indeterminate form, it's not equal to 0.
So we have to work our way around it some how.
 zepdrix
Notice that the denominator is the difference of squares,\[\large\rm x^21=x^21^2\]We have a rule for factoring the difference of squares, do you remember it? :)
 anonymous
So we have to multiply both top and bottom by x+1?
 zepdrix
Umm sure, that's another option!
That's pretty much the same thing I was doing, but in reverse.
 anonymous
Oh okay. That's how my teacher taught me. So after reducing or crossing out you would get 1/(x+1)?
 anonymous
And then you would have to plug in 1 right?
 zepdrix
Yes :)
 anonymous
Oh, that's what I forgot, been a couple years! :P Thank you again, Zepdrix.
 anonymous
Ohhhh that's why the answer is 1/2!
 zepdrix
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.
 anonymous
Okay I understood that. Thanks! Can you help me with number 7?
 zepdrix
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.
 zepdrix
Plug 3 directly into the function.
We get something like this,\[\large\rm e^{3}\cos\left(\frac{\pi(3)}{3}\right)\]yes?
 zepdrix
And now simplify :)
 anonymous
\[e^3\cos(3.14) = e^3(1) = e^3\]
 zepdrix
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
 anonymous
Lol sorry! But thank you soo much!! You are really good at helping others! :D
 zepdrix
yay team \c:/ we did it!
 anonymous
Thanks @Mikeyy1992! You were very helpful too! :)
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