Solving the Limiting Equation: x^3/e^(x^3) - 1 = 1

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Discussion Overview

The discussion revolves around evaluating the limit of the expression \( \frac{x^3}{e^{x^3}} - 1 \) as \( x \) approaches 0. Participants explore various methods to tackle this limit, including L'Hôpital's rule and Taylor series expansion, while also considering the relationship to a known basic limit involving the exponential function.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • James initially expresses confusion about how to approach the limit, noting that substituting \( x = 0 \) leads to an undefined \( \frac{0}{0} \) form.
  • One participant suggests using L'Hôpital's rule, which James admits he is unfamiliar with.
  • James proposes a substitution \( u = x^3 \) and draws a connection to the limit \( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \), suggesting that this might help in evaluating his limit.
  • Another participant cautions James about the implications of his substitution, emphasizing that \( u \) is dependent on \( x \) and that care must be taken when applying the reciprocal limit property.
  • A later reply clarifies that the Change of Variables Theorem can be applied, noting that while \( f(u) = \frac{e^u - 1}{u} \) is not continuous at 0, the conditions of the theorem can still be satisfied.
  • Elucidus explains how to derive the desired limit from the known limit involving \( e^u - 1 \) and the substitution, but does not assert that this is the only method to reach the conclusion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to evaluate the limit, with multiple approaches and some uncertainty about the implications of substitutions and theorems discussed.

Contextual Notes

There are unresolved questions about the application of L'Hôpital's rule and the continuity of functions involved in the limit evaluation. The discussion also highlights the potential pitfalls of variable substitution in limit problems.

james.farrow
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Hello everyone

I have a question thus:-

Prove

lim x -> 0 x^3/e^(x^3) -1 = 1

I have tried to tackle it as the previous questions but this has e in it whereas the others didn't.
If I put x = 0 we have 0/0 which is undefined.

I just don't know how to start...?

Many Thanks

James
 
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Do you know L'Hopital's rule?
 
Err unfortunately not. This is new to me and doesn't make much sense at the mo. I need to be able to work it out without a calculator...?

Not sure what to do at all
 
james.farrow said:
Hello everyone

I have a question thus:-

Prove

lim x -> 0 x^3/e^(x^3) -1 = 1

I have tried to tackle it as the previous questions but this has e in it whereas the others didn't.
If I put x = 0 we have 0/0 which is undefined.

I just don't know how to start...?

Many Thanks

James

Hi James, remember to use brackets where they're needed.

The lim x -> 0, x^3/e^(x^3) -1 is not equal 1, as written it's actually equal to -1.

However lim x -> 0, x^3/(e^(x^3) -1) = 1.

It would be best if you know either l'hospital's rule or Taylor series expansion to do this one.
 
Last edited:
Thanks for the help!

I'm not sure if I may have 'stumbled' across something, we are given a basic limit

lim x -> 0 (e^x -1)/x =1

Now this looks remarkably close to my question. I'm thinking if I let u = x^3 then I have

lim x -> 0 u/(e^u -1) which is the reciprocal of the given basic limit. I'm also sure that I have read the following property somewhere

lim x -> c f(x) = L

Then

lim x -> c 1/f(x) = 1/L

So taking my theory further as the given limit is 1 and my function with the substitution of u = x^3 is the reciprocal then 1/1 =1 which 'fits', but how do I write the limit containing my substitution...?

Or is the above just total b*ll*cks...!

Many Thanks

James
 
james.farrow said:
I'm not sure if I may have 'stumbled' across something, we are given a basic limit

lim x -> 0 (e^x -1)/x =1

Now this looks remarkably close to my question. I'm thinking if I let u = x^3 then I have

I haven't followed you 100%, but I can tell you here that you need to be careful. Letting u = x^3, it LOOKS like the x disappears from the top of the equation, but u is in fact dependent on x. It's a function of x. If your goal is to use the reciprocal rule, I'm pretty sure c has to be constant with respect to x (yet u is not).
 
Shame - thought I was onto something...

Just haven't a clue
 
james.farrow said:
Thanks for the help!

I'm not sure if I may have 'stumbled' across something, we are given a basic limit

lim x -> 0 (e^x -1)/x =1

Now this looks remarkably close to my question. I'm thinking if I let u = x^3 then I have

lim x -> 0 u/(e^u -1) which is the reciprocal of the given basic limit. I'm also sure that I have read the following property somewhere

lim x -> c f(x) = L

Then

lim x -> c 1/f(x) = 1/L

So taking my theory further as the given limit is 1 and my function with the substitution of u = x^3 is the reciprocal then 1/1 =1 which 'fits', but how do I write the limit containing my substitution...?

Or is the above just total b*ll*cks...!

Many Thanks

James

You do not need L'Hospital's Rule or Taylor series is you've already been shown that [itex]\lim_{x\rightarrow 0}\frac{e^x-1}{x}=1[/itex].

You are correct that if [itex]\text{If }\lim_{x\rightarrow a}f(x) = L \neq 0,\text{ then }\lim_{x\rightarrow a}\frac{1}{f(x)}=\frac{1}{L}[/itex].

What you also need (and alluded to in your post) is the Change of Variables Theorem (or one of them at least):

Change of Variables

If [itex]\lim_{x\rightarrow a}g(x) = b \text{ and }\lim_{u\rightarrow b}f(u) = c[/itex]
then [itex]\lim_{x\rightarrow a}(f\circ g)(x)=c[/itex] provided either

(1) f is continuous at b, OR
(2) there exists an open interval containing a such that for all [itex]x \neq a[/itex] in the interval, [itex]g(x) \neq b[/itex].

This last theorem is rarely ever shown to calculus students (and should be) but is used with reckless abandon in examples and exercises.

What you have is [itex]f(u) = \frac{e^u-1}{u} \text{ and }g(x) = x^3[/itex]. Unfortunately f is not continuous at 0, but x3 is invertible throughout the real line so we've satisfied part (2) of the theorem and hence

[tex]\lim_{x\rightarrow 0}\frac{e^{x^3}-1}{x^3}=\lim_{u\rightarrow 0}\frac{e^u-1}{u}=1[/tex] (Note [itex]u\rightarrow 0 \text{ as } x\rightarrow 0[/itex]).

The result you seek can be derived from there through the reciprocal.

I hope this is helpful.

--Elucidus
 

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