Trick for evaluating limits by substituting in 1/n

[cuallito]

TL;DR Summary

Trick I vaguely remember from Calc 1

Hi, I remember some sort of method for evaluating limits from Calc 1 that involved substituting in 1/n for x and simplifying. Does that sound familiar to anyone? Sorry I know that's vague, but all I can really remember about it. I can't find it mentioned anywhere in Stewart nor online :/

Thank you...

 

[RPinPA]

Well, that would change a $\lim_{x \rightarrow \infty}$ to a $\lim_{n \rightarrow 0}$ or $\lim_{x \rightarrow 0}$ to $\lim_{n \rightarrow \infty}$. Perhaps there are cases where that's a useful thing to do, but it's not a trick I recall seeing off the top of my head.

 

[Mark44]

Summary: Trick I vaguely remember from Calc 1

Hi, I remember some sort of method for evaluating limits from Calc 1 that involved substituting in 1/n for x and simplifying. Does that sound familiar to anyone? Sorry I know that's vague, but all I can really remember about it. I can't find it mentioned anywhere in Stewart nor online :/

Thank you...

This substitution appears fairly often in calculus textbooks.
Given the limit $\lim_{n \to \infty}(1 + \frac 1 n)^n$, you can use the substitution x = 1/n, and work with the new limit $\lim_{x \to 0^+}(1 + x)^{1/x}$ and evaluate the new limit.

 

[shrub_broom]

I believe you are talking about the Heine Theorem.

 

[Mark44]

I believe you are talking about the Heine Theorem.

I don't think so.
I am not familiar with the Heine Theorem, but I did find online descriptions of the Heine-Borel Theorem and the Heine-Cantor Theorem, neither of which had anything to do with limits, as far as I could see.

Also, the OP mentioned that his question was about Calc 1, which focuses mostly on limits and differentiation.

 

[WWGD]

I vaguely remember it used to compute derivatives somewhat similar to what Mark44 wrote , maybe in terms of dense subsets, as in :

$f'(x)=Lim_{ n \rightarrow \infty} [ f(x+1/n)-f(x)] / (1/n)$

Let me try to remember more details.

 

[shrub_broom]

For every function limit $lim_{x -> x_0} f(x) = A$, if there is a series s.t. $lim_{n -> \infty} a_n = x_0$, then $lim_{n -> \infty} f(a_n) = A$. By the way, $lim_{x -> x_0} f(x) = A$ iff $forall~a_n~s.t.~ lim_{n -> \infty} a_n = x_0, lim_{n -> \infty} f(a_n) = A$. This is what I mentioned.