Limits using Rates of Growth

[Matthewhistorian]

For a question like this , I know how to solve the limit normally but I was wondering if a possible way of solving it would be to break it up into n/1 multiplied by (2^n)/(n^3+4) and since both functions are a faster growing function over a slower function the end result is still infinity. If that doesn't work can anyone explain why and if there's any examples where you can't just multiply two fast/slow functions to get a fast/slow function?

 

[Deleted member 4993]

For a question like this View attachment 31026, I know how to solve the limit normally but I was wondering if a possible way of solving it would be to break it up into n/1 multiplied by (2^n)/(n^3+4) and since both functions are a faster growing function over a slower function the end result is still infinity. If that doesn't work can anyone explain why and if there's any examples where you can't just multiply two fast/slow functions to get a fast/slow function?

Have you tried using L'Hopital's rule?

 

[Matthewhistorian]

Have tried using L'Hopital's rule?

I already know how to solve the limit I was wondering if my alternate method worked

 

[Dr.Peterson]

For a question like this View attachment 31026, I know how to solve the limit normally but I was wondering if a possible way of solving it would be to break it up into n/1 multiplied by (2^n)/(n^3+4) and since both functions are a faster growing function over a slower function the end result is still infinity. If that doesn't work can anyone explain why and if there's any examples where you can't just multiply two fast/slow functions to get a fast/slow function?

Rather than talk about "fast/slow", why not just say you are writing the limit as the product of two limits that are both infinite.

[math]\lim_{n\to\infty}\frac{n2^n}{n^3+4}=\lim_{n\to\infty}n\cdot\lim_{n\to\infty}\frac{2^n}{n^3+4}[/math]
That seems straightforward -- assuming you have learned a theorem that supports this. What theorems do you have about infinite limits?

 

[Matthewhistorian]

Rather than talk about "fast/slow", why not just say you are writing the limit as the product of two limits that are both infinite.

[math]\lim_{n\to\infty}\frac{n2^n}{n^3+4}=\lim_{n\to\infty}n\cdot\lim_{n\to\infty}\frac{2^n}{n^3+4}[/math]
That seems straightforward -- assuming you have learned a theorem that supports this. What theorems do you have about infinite limits?

Thank you, I was using fast/slow because that's the terminology my calc teacher uses but I wasn't aware of the properties of rewriting limits of multiplications of each other thanks

 

[Dr.Peterson]

Thank you, I was using fast/slow because that's the terminology my calc teacher uses but I wasn't aware of the properties of rewriting limits of multiplications of each other thanks

It sounds like you must be taking an informal approach to calculus, without learning actual theorems on the basis of which you can be sure of your results.

This page discusses infinite limits properly, though its last set of facts stops just short of the particular theorem you need, for the product of two infinite limits (and it technically doesn't say anything about limits at infinity):

Calculus I - Infinite Limits

In this section we will look at limits that have a value of infinity or negative infinity. We’ll also take a brief look at vertical asymptotes.

tutorial.math.lamar.edu