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

[james.farrow]

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

 

[daniel_i_l]

Do you know L'Hopital's rule?

 

[james.farrow]

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

 

[uart]

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'hopitals rule or Taylor series expansion to do this one.

 

[james.farrow]

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

 

[Tac-Tics]

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).

 

[james.farrow]

Shame - thought I was onto something...

Just haven't a clue

 

[Elucidus]

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 \lim_{x\rightarrow 0}\frac{e^x-1}{x}=1.

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

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

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

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 f(u) = \frac{e^u-1}{u} \text{ and }g(x) = x^3. 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

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

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

I hope this is helpful.

--Elucidus