- #1

Elvz2593

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## Homework Statement

This is a problem I came up with when I was doing something similar in Spivak's Calculus; although a simpler version.

Suppose, we have [tex] f(x)=x^3 [/tex] and [tex]g(x)=x^2 [/tex]

find [tex]\lim_{x\rightarrow \infty} f(x)/g(x) [/tex]

## Homework Equations

N/A

## The Attempt at a Solution

So, I had 2 solutions for it. One says the limit exist, the other one says otherwise,

1. first method came from Spivak's book.

[tex]\lim_{x\rightarrow \infty} f(x)/g(x) = \lim_{x\rightarrow \infty} x^3/x^2 = \lim_{x\rightarrow \infty} (x^3/x^3)/(x^2/x^3) = \lim_{x\rightarrow \infty} 1/(1/x) [/tex]

let [tex] h(x)=1 [/tex] and [tex] j(x)=1/x [/tex]

[tex]\lim_{x\rightarrow \infty} f(x)/g(x) = \lim_{x\rightarrow \infty} h(x)/j(x) [/tex]

[tex]\lim_{x\rightarrow \infty} h(x)=1 [/tex]

[tex]\lim_{x\rightarrow \infty} j(x)=1/x=0 [/tex] this implies [tex] \lim_{x\rightarrow \infty} h(x)/j(x) [/tex] does not exist, otherwise [tex]\lim_{x\rightarrow \infty} h(x)= (\lim_{x\rightarrow \infty} h(x)/j(x) ) * ( \lim_{x\rightarrow \infty} j(x)) = 0 [/tex]

[tex]\lim_{x\rightarrow \infty} h(x) [/tex] can't both be 1 and 0.

2. the other method.

[tex] \lim_{x\rightarrow \infty} f(x)/g(x) = \lim_{x\rightarrow \infty} x^3/x^2 = \lim_{x\rightarrow \infty} x = \infty[/tex]

One of the solutions has to go. I probably just can't think straight since I have been up all night. Any help will be appreciated, my head is pretty much empty right now.