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 x
3 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