Something weird about limit at infinity?

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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.:confused:
 
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The first "method" is overly complicated but they are not contradictory. "[itex]\infty[/itex]" is not a number- saying that a limit is infinity is the same as saying the limit does not exist.
 
HallsofIvy said:
The first "method" is overly complicated but they are not contradictory. "[itex]\infty[/itex]" is not a number- saying that a limit is infinity is the same as saying the limit does not exist.
Saved my day, I can go to bed now.:zzz: