To What Extent Can You Manipulate Limits? (2024)

I'm currently taking Math Methods in Physics, and we're working on infinite series right now. In one of the examples, in order to find the limit, they do what you can see below in "Example 3"

As you can see, they turn the original equation into a ln equation, so that they can bring the exponent out in front, then change the sign so they can flip the ratio of the ln from "## 1/n ##" to "##n##". Then, they apply L'Hopital's rule and take the derivative to find the limit as that equation approaches infinity to be 0. Then, since they applied ln to the original function, they apply that value as an exponential value of e, which gives ## e^0=1##. That is an awful lot of changing an equation in multiple ways to get it where you want it to be, and I was surprised you can manipulate it to that extent, and still get the correct answer in the end.

So, what I'm wondering is if there is any limit as to how you can manipulate an equation like this to find the limit, so long as you apply the reverse of what you did to the end answer like they did? Can we do the same thing with any exponent e or logarithm to any limit? I mean I know adding certain things like multiplication or division and just reversing it at the end would deviate you from the correct answer (at least I believe it would), so I'm just trying to learn to what extent can we add things to the equation and just reverse it later like this, yet still get the correct limit in the end?

Ascendant0 said:

So, what I'm wondering is if there is any limit as to how you can manipulate an equation like this to find the limit, so long as you apply the reverse of what you did to the end answer like they did? Can we do the same thing with any exponent e or logarithm to any limit?

What you're asking about is the limit of a composite function and can you interchange the limit process and the outer function.
If ##\lim g(x) = a## and f is continuous at a, then ##\lim f(g(x)) = f[\lim g(x)]##.

Mark44 said:

What you're asking about is the limit of a composite function and can you interchange the limit process and the outer function.
If ##\lim g(x) = a## and f is continuous at a, then ##\lim f(g(x)) = f[\lim g(x)]##.

Thank you for that information. So from what I'm understanding from what you explained, I can manipulate g(x) however I want - ##ln(g(x))##, ##e^{g(x)}##, multiply x ##g(x)##, divide##/g(x)##, so long as I do it to g(x) in its entirety, and then at the end, make sure to "undo" whatever it is I did to it?

Now to clarify, this would only include things like multiplication, division, ln, and exponential, right? For example, I couldn't take ## 5 + g(x) ##, then manipulate that answer, then later at the end subtract "##g(x)-5##", as that would give an incorrect answer, am I right? Trying to make sure I understand exactly what I can and can't do for limits like this in the future.

Mark44 said:

Your example in a little more detail:
Let ##y = (\frac 1 n)^{1/n}##
Then ##\ln y = \ln (\frac 1 n)^{1/n}) = \frac {-\ln n} n##
Exponentiation both sides gives ##e^{\ln y} = e^{\frac {-\ln n} n}##
Notice that ##e^{\ln y} = y##,
so ##\lim_{n \to \infty} y= \lim_{n \to \infty} e^{\frac {-\ln n} n}##
## = e^{\left( \lim_{n \to \infty} \frac {-\ln n} n \right)}##
##= e^0 = 1##

I just read this after I posted. So from what I gather then, I can add or subtract as well then, as:

##y = (\frac 1 n)^{1/n}## would become ##y + 5 = (\frac 1 n)^{1/n} + 5##

And then before I take the limit, I'd just need to subtract the 5 again, and the answer would still be correct, right? Of course I know adding 5 to this problem is pointless, but I'm just trying to make sure that if in some other problem, addition (maybe to combine with a rational function) would in some way be useful, we can do that too, right?

The basis of this technique is the following property of a continuous function. If ##x_n \to x##, then ##f(x_n) \to f(x)##.

Note that in general the converse does not hold. So, you have be careful in cases where ##x_n## may not converge.

What you're using here is that if L is the limit, and ln L= a, then ##e^{ln L}=L= e^a ##

Ascendant0 said:

I just read this after I posted. So from what I gather then, I can add or subtract as well then, as:

##y = (\frac 1 n)^{1/n}## would become ##y + 5 = (\frac 1 n)^{1/n} + 5##

And then before I take the limit, I'd just need to subtract the 5 again, and the answer would still be correct, right? Of course I know adding 5 to this problem is pointless, but I'm just trying to make sure that if in some other problem, addition (maybe to combine with a rational function) would in some way be useful, we can do that too, right?

What you'd be using here is that , given ##f(x), g(x)##, if both the limits exist at a given point, ##Lim_{x\rightarrow x_0}f(x)+g(x)## is the sum of the limits.

Ascendant0 said:

So from what I gather then, I can add or subtract as well then

Yes, of course. Since you seem to be unsure of these, you should familiarize yourself with the properties of limits. The linked web page lists them: https://tutorial.math.lamar.edu/Classes/CalcI/LimitsProperties.aspx