Question

Given that f(x) is 2x^2/5x^2-9x-2, a) find the limit of f(x) as x approaches infinity
b) state the domain of the function

Answer 1

Well, there is a sort of trick to finding these rational fraction limits fast. But the drawn outway is totake the largest power of x in the denominator and divide every term in the problem by that power of x. So in your case, the largest power in the denominator is x^2. So you would then take x^2 and divide all terms in the problem by x^2.

Answer 2

and then?

Answer 3

but how would the denominator turn out if you divide by x^2? because i get the 2/5 part but what about the 9x-2?

Answer 4

Right, you would have -9/x and -2/x^2.
Well if you took those fractions separately to infinity what would you get?

Answer 5

I don't know :(

Answer 6

Well let's think of it this way. If I give you:
\[\frac{ 1 }{ x }\] now this is an example of when the power of x is higher in the denominator than in the numerator. So let's increasing raise the value of x in this fraction:
\[\frac{ 1 }{ 10 }->\frac{ 1 }{ 100 }->\frac{ 1 }{ 1000000000 }\]Can you see what the results are approaching if I were to continue?

Answer 7

It would get closer and closer to 0?

Answer 8

Correct. So thisis exactly what happens with the -9x and the -2. Because they have higher powers of x in the bottom, they go to 0. So if all of those go to 0, you're only left with 2/5, which is the answer.

Answer 9

thank you! How do I find the domain then?

Answer 10

The normal way you'd find a domain of a function :P

Answer 11

ahah (-infinity, 0}?

Answer 12

And then the (0,infinity)

Answer 13

Well, you need to factor the bottom xD

Answer 14

Set the denominator = to 0 and factor it to solve for x.

Answer 15

You know how to factor 5x^2-9x-2, correct?

Answer 16

lol no. I don't know how to do it when there's a number in front of the x^2

Answer 17

Ah. Yeah, I try and see the difficulty of the problem you're doing and sometimes I make incorrect assumptions based on it xD Okay, when the number in front of x^2 is not 1, we just have extra steps. You still need to find factors of (5)(-2) that add up to -9.

Answer 18

so -5, -1, -2, and 1? no that doesn't work. umm..

Answer 19

Well, 5 and -2 is -10. Factors of -10 that add up to -9 would be -10 and +1

Answer 20

so x+10 and x-1 would be the factors. and then set that equal to 0?

Answer 21

No, -10 and + 1. But there's more work than that when we have that number in front of x^2 to not be 1. What we do is replace the middle term with our factors. So we would rewrite out function as:
\[5x ^{2}-10x+x-2\]
Then we need to do factoring by grouping.

Answer 22

(x-2) and (x-2)?

Answer 23

bring the 5x out in front

Answer 24

so 5x (x-2) and (x-2)?

Answer 25

Well, you have to factor something out of each parenthesis. Its just in casethe only thing you can factor out is a 1. So we would have:
5x(x-2)+1(x-2), which means our factors are
(5x+1)(x-2)

Answer 26

and thenn set it equal to 0!

Answer 27

Each one, yes.

Answer 28

So as you can see, this means we have to exclude -1/5 and 2 from our domain. Do you know how to write that in interval notation?

Answer 29

(-infinity, -1/5} U {2, infinity} ?

Answer 30

Not quite. you left out an interval :3

Answer 31

We still have to include numbers in between -1/5 and 2. Those numbers still do exist.

Answer 32

so.. is that an intersection? [1/5, 2]?

Answer 33

We still use parenthesis for that portion.

Answer 34

\[(-\infty,-\frac{ 1 }{ 5 })\cup(-\frac{ 1 }{ 5 },2)\cup(2,\infty)\]

Answer 35

thank you so much! I actually learned how to do several concepts with your help. Thank you!

Answer 36

Yep, np. If you got other problems to look at while we're here, then go for it xD

Answer 37

I have one last question.. It's another domain question but more difficult in my opinion.

Answer 38

Go for it.

Answer 39

Find the domain. Use interval notation.
f o g, where f(x) = 1/(x+2) and g(x) = 4/(x-1)

Answer 40

Well, the first thing we need is the domain of f and the domain of g before we do the fog part.

Answer 41

It is already factored so I can just set each to 0?

Answer 42

so x = -2 amd x = 1

Answer 43

Right. So we need to exclude those from the domain. Now we do the fog part. Do you have any idea how to go about doing that?

Answer 44

plug in f into g? somehow..

Answer 45

g into f.

Answer 46

So that means the entire function of g(x) mustgo into every x that is in f(x).

Answer 47

g(x) = 4/0 though. Isn't that undefined?

Answer 48

It equals 4/(x-1), just like you wrote. Now you plug that into the x in f(x)

Answer 49

so 1/ (4/x-1) +2

Answer 50

would that change into 1 (x-1)/ 4+2?

Answer 51

Correct. Now you need to simplify it best you can. Well, I'm guessing they want you to at least.

Answer 52

No, it wouldnt. Not that easy unfortunately x_x

Answer 53

can i multiply the whole denominator to the top and cancel out the denominator then?

Answer 54

You could do a few things, whichever seems best to do. You could literally do the division ny flipping and multiplying like:
\[\frac{ 1 }{ \frac{ 4 }{ x-1 } +2}=\frac{ 1 }{ 1 }*(\frac{ x-1 }{ 4 }+\frac{ 1 }{ 2 })\]

Answer 55

so then x-1/4 + 2/4 = x-1+2/4 which equals x+1/4?

Answer 56

No, looks like you did some simplification wrong. What I would try to do is multiply top and bottom by x-1.

Answer 57

um so x-1/4 + 2x-2?

Answer 58

Correct. Now just simplify that a slight bit :P

Answer 59

can i factor out a (x-1) or a 2 anywhere?

Answer 60

Well, the 4 - 2 makes it 2+2x, meaning you can make the bottom 2(1+x)

Answer 61

then should I multiply the top by 2?

Answer 62

No need, you can leave it as is. Now that we finally have fog, what must we exclude from the domain?

Answer 63

the -2 and 1 from one of the beginning steps.

Answer 64

Yeah, but what about our new function, fog? We have to find numbers from that to exclude, too

Answer 65

(- infinity, -2) U (-2, 1) U (1, infinity) ?

Answer 66

You're forgetting fog that we just found xD

Answer 67

so I plug in the -2 and 1 to the x spots which gives me 3/2 and 0

Answer 68

No, you're overthinking it. Just look ONLY at what we found for fog. We have (x-1)/(2(1+x)) What must we exclude from the domain of that?

Answer 69

1

Answer 70

Nope, we only look at the denominator.

Answer 71

the two? or the (1+x)?

Answer 72

The 2 doesnt matter really, only the 1+x part.

Answer 73

so we have to take that out of the domain. but how?

Answer 74

How did we find the domain at the beginning?

Answer 75

we factored and then set it equal to 0.

Answer 76

Right. So set 2(x+1) = 0 and get x

Answer 77

x = -1

Answer 78

(- infinity, -1] U [-1, infinity)?

Answer 79

Right, that's what I was trying to get you to find xD So we have to exclude everything we found. So we have to exclude -2, -1, and 1.

Answer 80

oh great. um (-infinity, -2) U (-2, 0) U [1, infinity}?

Answer 81

All parenthesis, no brackets. And we never excluded 0, we can have 0. This is our interval and see if it makes sense:
\[(-\infty,-2)\cup(-2,-1)\cup(-1,1)\cup(1,\infty)\]

Answer 82

YAY, AWESOME! THANK YOU SO MUCH! Honestly learned much more than I did with my pre-calc teacher last year. Thank you!

Answer 83

Lol, np. Surprised you had that limit problem, though. People learn that in regular calculus xD

Answer 84

this was our summer packet to enter into AP Calc AB.. sort of a struggle after the whole summer, reallly.

Answer 85

Ah, gotcha. Well, limits are the very first thing you do in calculus, so if you can preview them, thats great. After limits is derivatives. So if you got those to look at it might help you out a ton to go over some of that stuff :P

Answer 86

Thank you so much again! :)

Answer 87

Yep, np :3