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

[Chronogallant]

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 I am not sure if its the right answer. pretty sure it isnt

 

[Ray Vickson]

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 I am not sure if its the right answer. pretty sure it isnt

You wrote
$$\lim_{x \to \infty} 5 + \frac{6x^2}{\sqrt{x^3}+ 2x^2+1}$$
If you really mean this, OK, it can stand as written. However, if you mean
$$\lim_{x \to \infty} \frac{5 + 6x^2}{\sqrt{x^3}+ 2x^2+1}$$
then you absolutely MUST use parentheses, like this:
(5+6x^2)/(sqrt(x^3) + 2x^2 +1).

 

[Chronogallant]

edited* whoops, thanks

 

[Mark44]

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 I am 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.

 

[Chronogallant]

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

 

[HallsofIvy]

Yes, that is correct. Roughly, the "dominating term" (the term with the highest power) in the both numerator and denominator is the $x^2$ term and the ratio of their coefficients is 6/2= 3.