Calculus 1 Homework problem - Find limit without L'Hospital's Rule

  • #1

Homework Statement



Lim ( 5+6x2)/(√(x3)) + 2x2 +1)
x->∞

Homework Equations



not allowed to use lhopitals rule

The Attempt at a Solution



first, i divided by x2, which yielded

(5/(x2) + 6 + √(x)) / (2 + 1/x2 )

then i assumed that thelim x--> infi of 5/x2 = 0, lim x--> infi 6 = 6, lim x---> infi sqrt(x) = infinity, lim x--- > infi 2 = 2, lim x---> infi 1/x2 = 0

so basically i ended up with the limit = to (6(∞))/(2) which im not sure if its the right answer. pretty sure it isnt
 
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Answers and Replies

  • #2
Ray Vickson
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Homework Statement



Lim 5+6x^2/(sqrt(x^3) + 2x^2 +1)
x->∞

Homework Equations



not allowed to use lhopitals rule

The Attempt at a Solution



first, i divided by x^2, which yielded

(5/(x^2) + 6 + sqrt(x)) / (2 + 1/x^2 )

then i assumed that thelim x--> infi of 5/x^2 = 0, lim x--> infi 6 = 6, lim x---> infi sqrt(x) = infinity, lim x--- > infi 2 = 2, lim x---> infi 1/x^2 = 0

so basically i ended up with the limit = to (6(∞))/(2) which im not sure if its the right answer. pretty sure it isnt
You wrote
[tex] \lim_{x \to \infty} 5 + \frac{6x^2}{\sqrt{x^3}+ 2x^2+1}[/tex]
If you really mean this, OK, it can stand as written. However, if you mean
[tex] \lim_{x \to \infty} \frac{5 + 6x^2}{\sqrt{x^3}+ 2x^2+1}[/tex]
then you absolutely MUST use parentheses, like this:
(5+6x^2)/(sqrt(x^3) + 2x^2 +1).
 
  • #4
34,156
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Homework Statement



Lim 5+6x^2/(sqrt(x^3) + 2x^2 +1)
x->∞
You need a pair of parentheses around the two terms in the numerator.

Homework Equations



not allowed to use lhopitals rule

The Attempt at a Solution



first, i divided by x^2, which yielded

(5/(x^2) + 6 + √(x)) / (2 + 1/x^2 )
This is a good approach, but you lost a term in the denominator. There should be three terms in the denominator, not two. BTW, √(x3) is the same as x√x.
then i assumed that thelim x--> infi of 5/x^2 = 0, lim x--> infi 6 = 6, lim x---> infi sqrt(x) = infinity, lim x--- > infi 2 = 2, lim x---> infi 1/x^2 = 0

so basically i ended up with the limit = to (6(∞))/(2) which im not sure if its the right answer. pretty sure it isnt
Correct, it's not the right answer. Your answer should not have ∞ in it unless the limit is actually infinity. Otherwise, we don't do arithmetic with infinity.
 
  • #5
oh gosh... so the missing term would be sqrt(x)/x, which would have a limit of 0 as it approaches infinity, due to the bottom scaling at a squared rate compared to the top? leaving the final limit as x approaches infinity as 3
 
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  • #6
HallsofIvy
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Yes, that is correct. Roughly, the "dominating term" (the term with the highest power) in the both numerator and denominator is the [itex]x^2[/itex] term and the ratio of their coefficients is 6/2= 3.
 

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