# Stuck on limits

1. Aug 11, 2009

### 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

2. Aug 11, 2009

### daniel_i_l

Do you know L'Hopital's rule?

3. Aug 11, 2009

### 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

4. Aug 11, 2009

### uart

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.

Last edited: Aug 11, 2009
5. Aug 11, 2009

### 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

6. Aug 11, 2009

### Tac-Tics

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).

7. Aug 11, 2009

### james.farrow

Shame - thought I was onto something...

Just haven't a clue

8. Aug 11, 2009

### Elucidus

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.