Asked by Maya
So I just have basic questions about limits which I just need to clear up. So when I'm solving for a limit , when the greater exponent is on the top of the fraction ex: x^2+2x/x+1 then the limit will always =infinity? And how do u know when a limit = 0 ? I just don't want to be confused on my exam tomrow, please help explain :)
Answers
Answered by
Steve
yes, the limit is always ∞.The limit is Zero when the greater power is in the bottom.
This is because when x gets really huge, on;y the highest power counts. You can throw all the lower degree stuff away.
So,
3x^3-7x+3
------------ -> 3x^3/x = 3x^2 -> ∞
x+4
If the powers are the same, then the limit is just the ratio of the coefficients:
3x^3-7x+3
--------------- -> 3x^3/6x^3 = 1/2
6x^3+9x^2-3x+2
More formally, you can divide top and bottom by the biggest power, and then since k/x -> 0 as x->∞, all the fractions with x in the bottom -> 0 and can be discarded.
3x^3-6x^2+4
----------------
6x^4-2x+7
Divide all by x^4 and you have
3/x - 6/x^2 + 4/x^4
-------------------------
6 - 2/x^3 + 7/x^4
now let x->∞ and you are left with
(3/x)/6 -> 0/6 = 0
This is because when x gets really huge, on;y the highest power counts. You can throw all the lower degree stuff away.
So,
3x^3-7x+3
------------ -> 3x^3/x = 3x^2 -> ∞
x+4
If the powers are the same, then the limit is just the ratio of the coefficients:
3x^3-7x+3
--------------- -> 3x^3/6x^3 = 1/2
6x^3+9x^2-3x+2
More formally, you can divide top and bottom by the biggest power, and then since k/x -> 0 as x->∞, all the fractions with x in the bottom -> 0 and can be discarded.
3x^3-6x^2+4
----------------
6x^4-2x+7
Divide all by x^4 and you have
3/x - 6/x^2 + 4/x^4
-------------------------
6 - 2/x^3 + 7/x^4
now let x->∞ and you are left with
(3/x)/6 -> 0/6 = 0
Answered by
Maya
So when the greater power is on top, the limit will equal infinity and when it's on the bottom then it will equal 0?
Answered by
Steve
Yes, I believe that is what I said in my first sentence.
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