Question

Find the Lim -> Inf. x^2+x+5 \ x^3 +1

Answer 1

Since the limit is approaching inf the answer would be \[\frac{ 0+0+5 }{ 0+1 } = \] Correct?

Answer 2

5. The answer i wrote was suppose to be 5

Answer 3

GHHH: Careful here. If you're taking the limit of this rational expression as x approaches infinity, you have no justification for replacing x with zero. Want to give this further thought?

Answer 4

Here is what I'd suggest:

1) Identify the highest power of x (it's x^3).
2) Divide EVERY term in numerator and denominator of this rational fraction by that highest power (x^3).
3) NOW, let x approach infinity. What's left?

Answer 5

\[\frac{ \frac{ 1 }{ x }+ \frac{ 2 }{ x } + \frac{ 5 }{ x }}{ \frac{ x ^{3}{x ^{3}} }{ x ^{3}{x ^{3}}+ \frac{ 1 }{ x ^{3} }}}\] = 0?

Answer 6

The second two X's on the bottom where suppose to be a fraction. I said zero because the top 3 turn into zeros becaus they're too large right? and the fractions are impossible to be found out?

Answer 7

You have x^3 + 1 in the den. and are to divide both terms by x^3, producing
1 + 1/(x^3), right?

I also suggested that you divide each term in the numerator by (x^3), which you have not yet done.

Mind re-doing this division? Once you're done, again try taking the limit as x approaches infinity.

Answer 8

Again, GHHH: Identify the HIGHEST POWER OF x (which is, in this case, x^3), and then divide EVERY TERM of the numerator and denom. by that highest power.

Answer 9

\[\frac{ \frac{ 1 }{ x } + \frac{ x }{ x } + \frac{ 5 }{ x }}{ \frac{ x ^{3} }{ x ^{3} } +\frac{ 1 }{ x ^{3} } } = \frac{ 0 + 0 + 0 }{ 1 + 0 } = \frac{ 0 }{ 1 } = 0? Right? \]

Answer 10

And I can't believe I missed the one in the dem, Sorry

Answer 11

Sorry, not zero. Undefined! ***

Answer 12

Please start over. ;)

Divide every term, every single one of those terms, by x^3.

Answer 13

Also, be certain that you're taking the limit as x approaches infinity (not zero).

Answer 14

I am doing terrible. Sorry, okay so here it goes. \[\frac{ \frac{ x^2 }{ x^3 } +\frac{ x }{ x^3 }+\frac{ 5 }{ x^3 }}{ \frac{ x^3 }{ x^3 }+ \frac{ 1 }{ x^3 } } \] = \[\frac{ \frac{ 1 }{ x } +\frac{ x }{ x^2 }+\frac{ 5 }{ x^3 }}{ 1 + \frac{ 1 }{ x^3 } }\] = (Have an inf sign on the bottom = (extremely close to) zero. So by that logic, all number in on the top turn into zero because numbers out of inf. (Same applies to the 1/inf in the dom). So leaving us with 0/1. Which is undefined.

Answer 15

Am I any closer?

Answer 16

first of all, you are correct in stating that you're left with 0/1; however, 0/1 is zero, not undefined.

Answer 17

Oh okay

Answer 18

I'm going to take your original expression and demonstrate how I'd divide every term in it by x^3:\[\frac{ \frac{ x ^{2}+x+5 }{ x ^{3} } }{ \frac{ x ^{3}+1 }{ x ^{3} } }\]

Answer 19

Okay

Answer 20

That gives us:
\[\frac{ \frac{ 1 }{ x }+\frac{ 1 }{ x ^{2} }+\frac{ 5 }{ x ^{3} } }{ 1 + \frac{ 1 }{ x ^{3}} }\]

Answer 21

Now, Ghost, if we let x approach infinity, the entire numerator becomes 0 and the entire denominator becomes 1. Please be absolutely certain you understand what's happening here; if you don't, ask questions.

Answer 22

so, if you end up with 0/1, that's = to 0, and 0 is the limit as x approaches infinity.

Answer 23

Just for clarification: Since the denominator of the numerators are X (inf) they are going to equal Zero because any part of inf = zero because the part is too small. Hence, numerator = zero. The denominator = 1 because x^3 \ X^3 = 1. the other some had the same principle as above applied to it to make it zero. Leaving us with 0/1 = zero.

Answer 24

If i do a 1 or 2 more, could you check it for me?

Answer 25

I'd be glad to help you with another problem.

Don't be offended, by my wanting to reword some of your "just for clarification" sentence for greater accuracy:

Answer 26

no, of course not. Better wording is welcome.

Answer 27

Clearer*

Answer 28

"Every fraction in the numerator goes to zero as x goes to infinity because each of these fractions has a constant numerator, whereas the denominator of each fraction increases without bound." There's no "too small" here.

Answer 29

All of the following go to zero as x increases to infinity:

Answer 30

\[\frac{ 1 }{ x }, \frac{ 1 }{ x^2 }, \frac{ 1 }{ x^3 },....\] is the general principle that comes into play here.

Answer 31

Try it: You can see for yourself that 1/x approaches 0 as x= 10, 100, 1000, etc.

Answer 32

AH okay. Quick question then, Will there be a case where inf is the numerator with a constant as a denominator?
As as for the try it, they increase by .0 each number. .1 -> .01 -> .001 -> etc.

Answer 33

To answer your first question: sure. This fraction would have no limit.
second question: actually, the fractions you've typed out here are DECREASING, which is my point exactly: as x goes to infinity, the fractions all get smaller, tending to 0 in the limit.