Question

how to solve
x-4/square root x -2

lim is getting close to 4

Answer 1

\[\large \lim_{x \rightarrow 4}\frac{x-4}{\sqrt x-2}\]Is that formatted correctly?

Answer 2

yes

Answer 3

With limits like this, we first want to try and get the square root out of the bottom. We'll do that by multiplying the top and bottom by the `conjugate` of the bottom term.\[\large \lim_{x \rightarrow 4}\frac{x-4}{\sqrt x-2}\left(\color{royalblue}{\frac{\sqrt x+2}{\sqrt x+2}}\right)\]
So on the bottom, do you remember what happens when we multiply conjugates? :)

Answer 4

wait, so if the square root is on the top, we will multiply by the conjugate of the top?

Answer 5

Yes, good call!
If our problem had looked like this,
\[\large \lim_{x \rightarrow 9}\frac{\sqrt x-3}{x-9}\]
We would have multiplied by the conjugate of the top.
\[\large \lim_{x \rightarrow 9}\frac{\sqrt x-3}{x-9}\left(\color{royalblue}{\frac{\sqrt x+3}{\sqrt x+3}}\right)\]

Answer 6

oh ok. So now we multiply out the top and leave the bottom in factored form

Answer 7

In the example I gave, yes.

But, In our problem, `x approaching 4`, we'll want to leave the top as multiplication. And we'll multiply out the bottoms to get rid of the square roots.

Answer 8

\[\large \lim_{x \rightarrow 4}\frac{(x-4)(\sqrt x+2)}{\color{orangered}{(\sqrt x-2)(\sqrt x+2)}}\]Understand how the orange terms will multiply out?
Here is what happens when we multiply conjugates:\[\large (a-b)(a+b)=a^2-b^2\]

Answer 9

oh ok . yes, I was told to only multiply the top and never the bottom though

Answer 10

Hmm interesting :o
You've probably been working with problems which had the square root in the top. :3

Answer 11

ok. so the the bottom cancel out, I can sub in 4 and get the answer

Answer 12

Bottom cancels with the (x-4) on top?
Yah sounds good! :)

Answer 13

could you help with another question

Answer 14

i'm stuck on part of it

Answer 15

sure, i can try :3

Answer 16

\[\sqrt{1-x} -1 /3x \]

Answer 17

lim x=0

Answer 18

I multiplied by \[\sqrt{1-x} +1 \]

Answer 19

\[\large \lim_{x \rightarrow 0}\frac{\sqrt{1-x}-1}{3x}\]Like that? :o

Answer 20

yes

Answer 21

Ok you did the correct multiplication, just getting stuck somewhere in the middle?

Answer 22

yea , I got \[x/3x (\sqrt{1-x}+1)\]

Answer 23

I think you accidentally dropped the negative from the x in the top. Looks good besides that though.

Answer 24

ummm, but when I sub in 0, I get 0 at the bottom and the answer is -1/6 . umm what negative?

Answer 25

See how we haven't cancelled anything out yet?
We've only done a little multiplication.

We need to be able to cancel something out before we can try subbing 0 in.
\[\large \frac{-\color{royalblue}x}{3\color{royalblue}x (\sqrt{1-x}+1)}\]
So that's where we're at right now, see any nice cancellations popping out at you? :)

Answer 26

the x, but i thought I couldn't do that since one has a 3 attached and one has not. also were did the negative come from the top

Answer 27

\[\large \left(\sqrt{1-x}+1\right)\left(\sqrt{1-x}-1\right) \quad = \quad 1-x-1 \quad = \quad -x\]See the small mistake you made on top?

Answer 28

I thought that if I multiplied the two negative x I would get a positive x though

Answer 29

The 3 is attached to the `Whole thing on the bottom`. Remember, multiplication is commutative, meaning we can apply multiplication in any order that we want.

Think of the 3 as being attached to the big other part instead.
Or think of it as a fraction that you can remove completely from the limit.
Whatever will help you disconnect the idea that the 3 is super glued to the x :) lol

\[\large \lim_{x \rightarrow 0}\;\large \frac{-x}{3x (\sqrt{1-x}+1)} \quad = \quad \lim_{x \rightarrow 0}\;\large \frac{1}{3}\cdot \frac{-x}{x (\sqrt{1-x}+1)}\]

Answer 30

\[\large \left(\sqrt{1-x}\right)^2 \quad = \quad (1-x)\]No the x won't change to a positive, it just eliminates the square root by squaring.

Answer 31

lol ok. so when we cancel out the x, the 3 will be beside the bracket and multiply through. oh ok. I get it now

Answer 32

We should be careful the way we use the words "cancel out" in this problem.
Since we're cancelling out everything from the top, it's important to remember that by saying "cancel out" we mean that the terms will divide to give us 1.

I'm only making the distinction because we're not left with 0 on top, we're left with a -1.

Answer 33

yes. Sorry, we use "cancel out" in class a lot and it stays with me. I know they divide out to 1

Answer 34

ok cool c:

Answer 35

i'm very sorry to ask, but I have another question that I worked out and got stuck

Answer 36

k whatchu got XD

Answer 37

\[\sqrt{3-x}-\sqrt{x-3}/x\]

Answer 38

limit =0

Answer 39

Hmm I'll bet we're running into the same issue that we had in the last problem ~ with the x's staying negative after we square the roots.

Answer 40

I multiplied by \[\sqrt{3-x}+\sqrt{x+3}\]

Answer 41

i ended up with \[1/x (\sqrt{3-x}+\sqrt{x+3})\]

Answer 42

\[\large \lim_{x \rightarrow 0} \frac{\sqrt{3-x}-\sqrt{x+3}}{x}\left(\color{royalblue}{\frac{\sqrt{3-x}+\sqrt{x+3}}{\sqrt{3-x}+\sqrt{x+3}}}\right)\]

Which becomes,\[\large \lim_{x \rightarrow 0} \frac{(3-x)-(x+3)}{x(\color{royalblue}{\sqrt{3-x}+\sqrt{x+3}})}\]

Answer 43

Which becomes,\[\large \lim_{x \rightarrow 0} \frac{-2}{\color{royalblue}{\sqrt{3-x}+\sqrt{x+3}}}\]

Answer 44

You got a 1 on top somehow? hmmm...

Answer 45

See how the squaring got rid of the roots, but it didn't change the sign on the x's?
After that, you have to remember to distribute the negative to each term in the second set of brackets - the orange one.

\[\large \lim_{x \rightarrow 0} \frac{(3-x)-(\color{orangered}{x+3})}{x(\sqrt{3-x}+\sqrt{x+3})}\]

Answer 46

oh ok. I'm just having trouble multiplying out the square roots . Will that be all that I can do after I get 2 on top

Answer 47

Distributing the negative gives us,\[\large \lim_{x \rightarrow 0} \frac{3-x\color{orangered}{-x-3}}{x(\sqrt{3-x}+\sqrt{x+3})}\]

Answer 48

Oh were you able to get 2 on the top?

Answer 49

no, how did you get it ? do I expand the top out?

Answer 50

Do you understand how I got to the last step? :o

Answer 51

yes, you multiply everything in the second bracket with the negative. But why did you remove the brackets

Answer 52

brackets don't mean anything if there is nothing being applied on the outside. That is one of the difficult things to understand if you really remember your PEMDOS from earlier math classes.

You'll go, "Hmm wait a sec, I thought I have to deal with things inside of brackets first!" No no no, only if they have something being applied on the outside should you really worry about that.

Example:\[\large (7+x)+(3+x) \quad = \quad 7+x+3+x\]If something was being applied to the outside of the brackets, we would have to deal with that first,\[\large (7+x)+2(3+x) \quad = \quad 7+x+(6+2x) \quad =\quad 7+x+6+2x\]

Answer 53

umm ok. But I always thought that if brackets were shown then it would mean I would have to expand everything out?

Answer 54

Expand what out? :|
If you have 2 sets of brackets being `multiplied` together, then yes we have to "expand" them out ~Multiply the brackets together to get all of the cross-terms in the middle and outsides.

We don't have that happening in this problem. We have 2 sets of brackets being `added`.

Answer 55

oh ok . so the top will be 3-x-x-3

Answer 56

ya

Answer 57

oh i understand how you got 2 now

Answer 58

-2*

Answer 59

-2 sorry, on the equation the top works out to 3-x-x-3 on the part we multiply it with it works out to 3-x+x+3 (Everything cancels out in this since the all equal 0, leaving 1 on the top) you end up with -x on the equation, you multiply them together you get negative -2x.

Answer 60

I'm not quite sure what you just said there.. lol XD
But yes the 3's should cancel out, leaving us with -2x on top.
And then you should see a nice cancellation from there.

Answer 61

the top will be 3-x-x+3 +3-x+x+3

Answer 62

no... im not sure where you're getting that from.

Answer 63

could you show me how to got -2 then?

Answer 64

\[\large \begin{align*}\left(\sqrt{3-x}-\sqrt{x+3}\right)&\left(\sqrt{3-x}+\sqrt{x+3}\right)\\ \\&=(\sqrt{3-x})^2-(\sqrt{x+3})^2\\ \\&=(3-x)-(x+3)\\ \\ &=3-x-x-3\\ \\&=-x-x\\ \\&=-2x\end{align*}\]

Answer 65

oh ok. So you square the parts that are similar since you have two of them, which will take away the square root, then you removed the brackets leaving you with -2x, the x can divide out

Answer 66

ya, good times.

Answer 67

lol, thank you so much, you helped so so much. thank you.

Answer 68

hehe c: np, haf a good night c:
i gotta finish studying for test tomorrow! :O

Answer 69

you too. night and good luck