Limit of sqrt(9x^6 - x)/(x^3 + 1) as x approaches infinity

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Limit of sqrt(9x^6 - x)/(x^3 + 1) as x approaches infinity. We divide the numerator and denominator by the sqruare root of x^6 to do this. Note sqrt(x^6) = x^3 in this problem because x is approaching infinity.

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A good trick for some problems is to plug in large numbers (10^3 e.g) and get an approximate answer. Plug in 10^4 and see if that approximate is reaching an asymptote. You can do that mentally for x->inf.
The actual solution is neat too.

nickhill

Thanks dear I struggled before with similar problems

jamesjohn

Since it’s a limit you can kinda break the rules of algebra

In the top term (9x^6-x)^1/2, the x is redundant at infinity due to the presence of a larger power so you can just evaluate the top term as (9x^6)1/2 which is just 3x^

For the bottom term where you have x^3+1, the 1 is infinitely small next to infinity, so the equation can be approximated to 3x^3/x^3 which is just three

Understanding how infinity works with limits can save you a whole lot of algebra

blackninja

Yes true if there is negative infinity, then there is little problem, easier with positive infinity!!thanku for this question sir

atuljiupadhyay

Hai everyone I am from Indonesia I love math because this is so beautiful nice to meet you

mathematics

Could you please let me know, from which book, you chose this question ?

I used L’Hopital’s rule and got zero.

hussainfawzer

once, just once I will raise up and start learning all thins stuff. Just for to be a human

magamagabovich

is this supposed to be hard or easy problem?

KAID

You should be putting equal signs in front of all of those lim's!

billtruttschel

If infinite change to negative, what's the answer?

VictorGonzalez-rwrt

Hi, I don't mean to hate but there is a much easier way of instantly knowing the answer to this problem. Using the understanding of powers to obtain the answer for limits approaching infinity, we can simply do the square root of the leading term in the numerator which becomes 3x^3. Because the leading numerator and denominator both have the same exponent, we simply take the ratio of the coefficients and get 3/1 which is just 3. I hope that didn't come off as pretentious but the method you used seems unnecessarily complicated.

potatolordozai

If the final value is square root of 9 over 1, then shouldn't the answer be +/-3?

niket

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