Question

Limit of (x^7-1)/(x^6+1) as x approaches infinity.

Answer 1

infinity

Answer 2

because 7>6

Answer 3

SO when the Degree of the numerator is bigger than the degree of the denominater. it's infinity???

Answer 4

If I follow that rule, isn't it also possible that the answer doesn't exist? how do you know which is which?

Answer 5

\[\lim _{x \rightarrow \infty} \frac{x^7-1}{x^6+1}=\frac{\infty}{\infty}\]

Answer 6

Why?

Answer 7

you have an inderterminate form so use l'hopitals

\[f'(x)=7x^6\]
\[g'(x)=6x^5\]
still infinity, take f'' , g''

\[f''(x)=42x^5\]
\[g''(x)=30x^4\]
eventually the bottom will disappear leaving you with a x on top which will get you inifinity

Answer 8

ok heres the work i got:
lim x--> o-o(infinity)
it is an "indeterminate form of o-o/o-o
so use the largest power denominator and multiply the denominator and numerator by it
so you'll get like:
lim x --> o-o x^7-1 * 1/x^6 / x^6+1 * 1/x^6
-Now remembering the reciprocal power rule here (if an infinity is over a real number it goes to 0) you're going to get x/1
So what is lim x--> o-o of x/1?

Answer 9

No there's an easier way to explain this..

Answer 10

Basically since the top is higher power... it approaches infinity faster than the bottom

Answer 11

ya, very true, but what if her teacher wants the work, and i think it helps to understand why

Answer 12

you can also use l'hopitals rule.
\[\lim_{x \rightarrow \infty} \frac{ f(x) }{ g(x) } = \lim_{x \rightarrow \infty} \frac{ f'(x) }{ g'(x) }\]

where f(x) = x^7-1 and g(x) = x^6 +1
f'(x) = 7x^6 and g'(x) = 6x^5.
\[\frac{ 7*x^6 }{ 6x^5 } = (7/6)*x = infinity\]

Answer 13

ohh

Answer 14

\[\lim _{x \rightarrow \infty} \frac{x^7-1}{x^6+1}=\lim _{x \rightarrow \infty} \frac{x^6}{x^6}\frac{x-1/x^6}{1+1/x^6}=...\]

Answer 15

l'hospitals is overkill for such a simple problem

Answer 16

eh, I think like a calculator, and a calculator would solve it wth l'hopitals

Answer 17

because it's much easier to program if (both = infnity) , then take derivatives..

Answer 18

Ohhh. Gotcha. Thank you. I like the I'hopital rule. I've never heard of it. Thank you.

Answer 19

Ok guys. how about helping me with one more?

Answer 20

you should be careful if you are not familiar with L'Hospitals rule. It is easy to get wrong answers if certain conditions are not met

Answer 21

\[\lim_{x \rightarrow \infty } x ^{3}-5x ^{2}\]

Answer 22

Notice...
\[x ^{3}-5x ^{2}=x^2(x-5)\]

Answer 23

yesss

Answer 24

as \(x\to\infty\) you have \(x^2\to\infty\) and \(x-5\to\infty\)

Answer 25

So the answer is infinity?

Answer 26

yes

Answer 27

and if it were aprouching negative infinity, would it be negative infinity?

Answer 28

in this case ...yes

Answer 29

Ok thanks! Eee, I just got helped by Data ^_^ lol

Answer 30

you are welcome