Determining the Rate at Which Functions approach Infinity

  • #1
Dopplershift
59
9
With basic fractions, the limits of 1/x as x approaches infinity or zero is easily determine:

For example,
\begin{equation}
\lim_{x\to\infty} \frac{1}{x} = 0
\end{equation}

\begin{equation}
\lim_{x\to 0} \frac{1}{x} = \infty
\end{equation}

But, we with a operation like ##\frac{f(x)}{g(x)}##, you have to determine which function approaches infinity faster. If f(x) approaches infinity faster than g(x) then the answer is infinity; likewise if g(x) approaches infinity faster, than the answer is zero.

Do we determine which functions go to infinity faster simply by L'Hospital's rule in which we keep taking derivatives until a constant appears either on the bottom or top.

For example if I have the following"

##\lim_{x\to\infty} \frac{x^{1000x}}{e^{0.001x}}## = ?

I can argue that the answer is zero, because after I take 1000 derivatives with L'Hospitals rule, the function on top because some constant, while the bottom remains an exponential function;

##\lim_{x\to\infty} \frac{c}{e^{0.001x}} = 0##


therefore, can I use that to prove that the function e^x goes to infinity faster than some function x^{cx}?

But with that logic, how can we compare the speeds of such functions as n! or e^(x)? Considering e^x never becomes a constant and n! is non-differentable?

This is just a question I have out of curiosity. Thanks!
 
Last edited by a moderator:

Answers and Replies

  • #2
pwsnafu
Science Advisor
1,082
85
Well, to begin with you want to compare ##e^n##, and not ##e^x##, to ##n!##.
As for the proof this is a standard problem in first year calculus (if said course contains sequences and series). Hint: you use sandwich/squeeze theorem.
 
  • #3
36,424
8,403
Dopplershift said:
For example,
...
\begin{equation}
\lim_{x\to 0} \frac{1}{x} = \infty
\end{equation}
The limit in this example does not exist.The left- and right-sided limits are not the same, so the two-sided limit does not exist.
 
  • #4
mathman
Science Advisor
8,077
547
With basic fractions, the limits of 1/x as x approaches infinity or zero is easily determine:

For example,
\begin{equation}
\lim_{x\to\infty} \frac{1}{x} = 0
\end{equation}

\begin{equation}
\lim_{x\to 0} \frac{1}{x} = \infty
\end{equation}

But, we with a operation like ##\frac{f(x)}{g(x)}##, you have to determine which function approaches infinity faster. If f(x) approaches infinity faster than g(x) then the answer is infinity; likewise if g(x) approaches infinity faster, than the answer is zero.

Do we determine which functions go to infinity faster simply by L'Hospital's rule in which we keep taking derivatives until a constant appears either on the bottom or top.

For example if I have the following"

##\lim_{x\to\infty} \frac{x^{1000x}}{e^{0.001x}}## = ?

I can argue that the answer is zero, because after I take 1000 derivatives with L'Hospitals rule, the function on top because some constant, while the bottom remains an exponential function;

##\lim_{x\to\infty} \frac{c}{e^{0.001x}} = 0##


therefore, can I use that to prove that the function e^x goes to infinity faster than some function x^{cx}?

But with that logic, how can we compare the speeds of such functions as n! or e^(x)? Considering e^x never becomes a constant and n! is non-differentable?

This is just a question I have out of curiosity. Thanks!

I can argue that the answer is zero, because after I take 1000 derivatives with L'Hospitals rule, the function on top because some constant, while the bottom remains an exponential function;
Incorrect! [itex]x^{1000x}[/itex] has infinite non-zero derivatives! You are confusing it with [itex]x^{1000}[/itex].
 

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