Question

Medal will be awarded (Calculus)

Answer 1

Is there a problem?

Answer 2

Sorry, I can't help with that. But if you need history help, I can do that.

Answer 3

Ha thanks

Answer 4

You're welcome? Yeah, just send me a message if you need help with history.

Answer 5

For continuity, we need `connectedness` so to speak.
For differentiability, we need `smoothness`.

We're given a piece-wise function and we'll need to make sure that the pieces "connect".

Answer 6

So we'll use limits to achieve this :)

Answer 7

The limit from the left must agree with the limit from the right.

Answer 8

\[\Large\rm \lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)\]So we need to check this :)

Answer 9

If we're approaching 1 from the left side, which PIECE are we dealing with?

Answer 10

5x? o.o

Answer 11

Sorry I'm confused by what you mean by "piece"

Answer 12

The top piece: 2x+3, we use that when x is LESS THAN 1, ya? :)
So in other words, we use that piece when we're on the `left side of 1`, make sense?

Answer 13

Oohh Yes!

Answer 14

\[\Large\rm \lim_{x\to1^-}2x+3=\lim_{x\to1^+}f(x)\]So that's what we're using when we approach x from the left side. That's what our function looks like over there.
How bout on the other side? :3 pretty obvious, it's just going to be the other piece, ya?\[\Large\rm \lim_{x\to1^-}2x+3=\lim_{x\to1^+}5x\]

Answer 15

From there, you don't have to do anything fancy.
Just plug in x=1 for both sides,
and find out if there is continuity.
You should end up with a true statement if there is.

Answer 16

Ohhhh so it's 5?

Answer 17

No, not just 5.
But you end up with 5=5, yes?
Which is a true statement, so the pieces DO connect.
So yes, we have continuity! :)

Answer 18

Oooooooooooooh okay :)

Answer 19

yay team \c:/

Answer 20

-high fives-

Answer 21

Do you mind helping me with a few other problems? Just two more <.<

Answer 22

k

Answer 23

Thanks :D

Answer 24

Okay \[\lim_{x \rightarrow 0} \frac{ \sqrt{x+5}-x \sqrt{5}}{ x}\]

Answer 25

With limits, you have to learn all these weird little tricks.
This one involves `multiplying by a conjugate`.

Conjugates look like this: \(\Large\rm a-b,\quad a+b\) they look the same, but the sign between them is different.
When you multiply them, you get the different of their squares.\[\Large\rm (a-b)(a+b)=a^2-b^2\]Does this look familiar? :)

Answer 26

No

Answer 27

:(

Answer 28

Example:\[\Large\rm (7-3x)(7+3x)=7^2-(3x)^2=49-9x^2\]

Answer 29

With our problem it's going to look a little different, since we have square roots involved.

Answer 30

Example:\[\Large\rm (\sqrt{x+1}-\sqrt x)(\sqrt{x+1}+\sqrt x)=\sqrt{x+1}^2-\sqrt x^2\]\[\Large\rm =x+1-x\]

Answer 31

That's what going to happen for us. We'll square some square roots and some stuff will happen.
Let's see how it looks :)
We'll multiply both the top and bottom by the `conjugate of the top`.

Answer 32

\[\Large\rm \lim_{x\to0} \frac{\sqrt{x+5}-x\sqrt{5}}{x}\color{royalblue}{\left(\frac{\sqrt{x+5}+x\sqrt{5}}{\sqrt{x+5}+x\sqrt{5}}\right)}\]

Answer 33

Don't multiply out the bottoms! Leave them alone.\[\Large\rm \lim_{x\to0}\frac{?}{x(\sqrt{x+5}+x \sqrt 5)}\]What do we get in the top though? :)
What do you think?

Answer 34

I know the beginning of it is x but the two square roots aren't like terms..

Answer 35

\[\Large\rm \lim_{x\to0}\frac{\sqrt{x+5}^2-(x\sqrt5)^2}{x(\sqrt{x+5}+x \sqrt 5)}\]We get squre minus square, ya?

Answer 36

Ohh okay

Answer 37

\[\Large\rm =\lim_{x\to0}\frac{x+5-5x^2}{x(\sqrt{x+5}+x \sqrt 5)}\]Hmm did you post the question correctly? 0_o it doesn't look like this is going to work out nicely.

Answer 38

Was the x supposed to be under the root with the 5 in the second term?

Answer 39

I'll screen shot it

Answer 40

http://gyazo.com/c0dbe5817a84fb20cc9ad67023703ed3

Answer 41

^equation

Answer 42

Ya you put an extra x in there silly! >.<

Answer 43

Oh goodness I'm sorry Dx

Answer 44

\[\Large\rm \lim_{x\to0} \frac{\sqrt{x+5}-\color{red}{x}\sqrt{5}}{x}\color{royalblue}{\left(\frac{\sqrt{x+5}+\color{red}{x}\sqrt{5}}{\sqrt{x+5}+\color{red}{x}\sqrt{5}}\right)}\]Ok so these red x's were not supposed to be in the problem.

Answer 45

That will clean things up nicely.
Instead we'll have:\[\Large\rm \lim_{x\to0}\frac{\sqrt{x+5}^2-\sqrt5^2}{x(\sqrt{x+5}+x \sqrt 5)}=\lim_{x\to0}\frac{x+5-5}{x(\sqrt{x+5}+x \sqrt 5)}\]

Answer 46

You can combine the 5's ya?\[\Large\rm =\lim_{x\to0}\frac{x}{x(\sqrt{x+5}+x \sqrt 5)}\]What other simplification can we do? :)

Answer 47

take the x out? o.o

Answer 48

Divide the x's? Ya seems like a good idea.\[\Large\rm =\lim_{x\to0}\frac{\cancel{x}}{\cancel{x}(\sqrt{x+5}+x \sqrt 5)}\]What does that leave you with in the top? :)
Don't you dare say 0, lol

Answer 49

1 o.o

Answer 50

\[\Large\rm =\lim_{x\to0}\frac{1}{(\sqrt{x+5}+x \sqrt 5)}\]Ok good.

Answer 51

Maybe I forgot to mention at the start..
we're ultimately just trying to plug in x=0.

In the original form, x=0 gave us a big problem, a 0 in the denominator.
We go through some process, and try to cancel something out, and then we try again.

Answer 52

So if we plug in x=0 from this point, do we run into a problem still? :o

Answer 53

\[\frac{ 1 }{ 2\sqrt{5} }\] ?

Answer 54

Oh oh oh sorry I was still using the old one. I forgot to remove that extra x again.\[\Large\rm =\lim_{x\to0}\frac{1}{(\sqrt{x+5}+\sqrt 5)}\]But yessss, good job! :)

Answer 55

So that's the value of our limit.

Answer 56

Oh.. so that was your second account? o.o

Answer 57

What? 0_o
Oooo crap.

Answer 58

Bustedddd >.<

Answer 59

-Is confused- ._.

Answer 60

Next question? c:

Answer 61

So it is you!! :O

Answer 62

Anywho, next question \[\frac{ \left| 3-x \right| }{ 9-x^{2} }\]

Answer 63

Oh crap

Answer 64

\[\lim_{x \rightarrow 3+}\] then the equation I just put down :I

Answer 65

\[\Large\rm \lim_{x\to3^+}\frac{|3-x|}{9-x^2}\]Do you understand how to factor the denominator?

Answer 66

Yes

Answer 67

For the numerator, remember what absolute value function looks like?
It's like a V shape.

Answer 68

We define it this way:\[\Large\rm |x|=\cases{\quad x,& x$\gt$0\\ \rm -x,&x$\lt$0}\]

Answer 69

So for our problem,\[\Large\rm |3-x|=\cases{~~\quad 3-x, &x$\gt$3\\ \rm -(3-x), &x$\lt$3}\]

Answer 70

It's the left branch of the V shape when x is less than 3,
and it's the other one when x is greater than 3.

Answer 71

brb

Answer 72

\[\Large\rm \lim_{x\to3^+}\frac{|3-x|}{9-x^2}\]Our limit is from the RIGHT.
So we want the one that corresponds to x being greater than 3.

Answer 73

I think maybe I wrote those pieces backwarrrrds 0_o sec thinking

Answer 74

So there is the graph of the numerator, the absolute value function, shifted 3 to the right.