Question

Studying for an exam and I need help through a few problems, help?

Answer 1

Okay, I'll help you out.

Answer 2

Thank you!
The first thing I need some help with is calculating the limit algebraically

Answer 3

Okay so we have:
\[\lim_{x \rightarrow -4}\frac{ 3x ^{2}+10x-8 }{ x ^{2}-16 }\]
First off, let's look at it.
yeah, if I replace, i would have 0 on the denominator, and we couldn't go any further.
Let's see if we can factor it so it doesn't bother us much.
Aha! it's a square difference. Let's apply that:
\[\lim_{x \rightarrow -4}\frac{ 3x ^{2}+10x-8 }{ (x+4)(x-4) }\]
Good... Now... Let's see if we can factor that equation on the numerator...
How would you do that?

Answer 4

Factored out it would be (3x-2)(x+4) right?

Answer 5

Then (x+4) cancels out and just plug in 4 which gives me 7/4

Answer 6

Good!, so it'll look like this:
\[\lim_{x \rightarrow -4}\frac{ (3x-2)(x+4) }{ (x+4)(x-4) }\]
look at that! we can simplify the (x+4).
Ending up with:
\[\lim_{x \rightarrow -4}\frac{ (3x-2) }{ (x-4) }\]
If we apply the limite, it looks like this:
\[\frac{ 3(-4)-2 }{ (-4)-4 }=\frac{ -18 }{ -8 }=\frac{ 18 }{ 8 }\]
Voilá!

Answer 7

Whoops! I missed out that 3(-4)-2 = -14

@terenzreignz Thank you ^^"

Answer 8

@Naavidya was way ahead of me, with her answer: 7/4 though.

:)

Answer 9

Yeah, I was trying to figure out how to get past the denominator issue ^^*
The next one deals with rationalizing and I'm not sure how to start

Answer 10

@terenzreignz Can you help her? I'm very tired, it's 2:00 here :P

Answer 11

@Naavidya The keyword is 'conjugate' :)

Answer 12

Ever heard of it?

Answer 13

Is it the opposite of an equation or something like that?

Answer 14

Sort of. I just imagine it as 'changing the sign' between the 'normal part' and the 'weird part'

That's a very rough way to put it, but I only put it like that because you'll encounter the word conjugate again sometime in the future, maybe with complex numbers, etc, and its basically the same thing...

ANYWAY

What I like to call the 'weird part' here is the par
t inside the square root, namely, \(\Large \sqrt{7x-5}\) and the 'normal part' is the one outside, the \(\Large 4\) and they're separated by a minus sign.

Answer 15

Do we rationalize the function by multiplying the conjugate?

Answer 16

Multiplying both the numerator and denominator by the conjugate, yes :)
Keeping this in mind, of course:
\[\Large \left[\sqrt P +Q\right]\left[\sqrt P - Q\right]= P - Q^2\]

Answer 17

\[\frac{ x-3 }{ \sqrt{7x-5}-4 }\times \frac{ \sqrt{7x-5}+4 }{ \sqrt{7x-5}+4 }\]

Answer 18

Yes... very good :)

Answer 19

Don't stop now, you're doing great :D
Just simplify, using that thing I reminded you with (the P Q thing)

Answer 20

So multiplied out it should be \[\frac{ x \sqrt{7x-5}+4x-3\sqrt{7x-5}-12 }{ 7x-5-16+4\sqrt{7x-5}-4\sqrt{7x-5} }\]
And then cancel out?

Answer 21

I suggest you don't multiply the numerator through, leave it in its factored form ;)
\[\Large (x-3)\left(\sqrt{7x-5}+4\right)\]

Answer 22

But do simplify the denominator :)

Answer 23

So the denominator would be \[7x-21\] after simplifying?

Answer 24

That's good :)
\[\Large \frac{(x-3)\left(\sqrt{7x-5}+4\right)}{7x-21}\]

Answer 25

And to avoid the 0 in the denominator we factor out 7 and then cancel out the top and bottom?

Answer 26

\[\Large \frac{(x-3)\left(\sqrt{7x-5}+4\right)}{7(x-3)}\]
You seem eager :)
\[\Large \frac{\cancel{(x-3)}\left(\sqrt{7x-5}+4\right)}{7\cancel{(x-3)}}\]

Answer 27

I just catch on quickly :D
And just plug in 3?

Answer 28

If it doesn't result in anything ugly, such as 0/0, why not? ^_^

Answer 29

And that give 8/7 ?

Answer 30

Brilliant :)

Answer 31

The 3rd one kinda has me confused on what to even do

Answer 32

Why don't you simplify the numerator? :)

Answer 33

IE expand, and simplify

Answer 34

It actually kinda is :D

Answer 35

Oh, it just ends up being -6+t which equals -6 ^^*

Answer 36

Yup :)
And the last one... possibly the easiest or the hardest, depending on how you look at it...

Answer 37

Yeah, I'm clueless again...

Answer 38

Any gut feel?

Answer 39

I feel that the limit doesn't exist but I'm not sure how to explain

Answer 40

It doesn't :D
And here's why...
when x goes to 2 from the left, 2 > x
meaning
2-x > 0

Means |2-x| = 2-x
right?

Answer 41

Yeah, right

Answer 42

But... when x goes to 2 from the right, we get
2 < x
2-x < 0

|2-x| = -(2-x)

right?

Answer 43

Yeah, got it

Answer 44

So, from the left, our function is effectively
\[\Large \frac{2-x}{|2-x|}= \frac{2-x}{2-x}\] approaching 1, right?

Answer 45

Right

Answer 46

But from the right, since |2-x| = -(2-x) if x approaches from the right, we effectively get
\[\Large \frac{2-x}{|2-x|}= \frac{2-x}{-(2-x)}\]
approaching -1.

Answer 47

So it doesn't exist because they're not the same?

Answer 48

Yup :)
From the left, the function is (actually constantly) equal to 1
But from the right, the function is (also constantly) equal to -1

Answer 49

Since they don't 'collide' (lol) the limit doesn't exist XD

Answer 50

Okay, now I get why XD

Answer 51

@Naavidya so that about wraps it up, I hope?

Answer 52

Yup, thanks for the help :)

Answer 53

No problem.
You seem ready for that test :D