find the lim x->0 (e^(-x) -1) / (1-e^x)

Question

i did this using a table and the found the limit but is there a way to do this algebraically

Answer 1

You can do this (and mnore complicated problems) easily using Taylor expansions.

We have:

exp(x) = 1 + x + x^2/2 + x^3/6 + ...

Substituting this in the formula and exäding gives:

(e^(-x) -1) / (1-e^x) =

[-x + x^2/2 + ...]/[-x - x^2/2 + ...] =

[1 - x/2 + ...]/[1+x/2 + ...] =

1 - x + ....

Formulated rigorously, this means that in a neighborhood of x = 0, we have:

(e^(-x) -1) / (1-e^x) = 1 + O(x)

Where O(x) means that there exists a neighborhood of zero such that

|O(x)| < M |x|

for some constant M. See also here for general use of this notation:

http://en.wikipedia.org/wiki/Big_O_notation

This then means that the limit equals 1.

In this particular case, you could also have used L'Hôpital's rule, but that's equivalent to this derivation.

Answer 2

I would attempt l'Hôpital's rule first, as it is simpler and takes less time, an important factor for exams. However it depends on whether the course curriculum includes the rule or Taylor's expansion.

Using l'Hôpital's rule, when finding the limit of an apparently indefinite quantity involving a numerator and a denominator, we would replace the numberator and denominator by their respective derivatives and find the limit again. The rule can be applied repeatedly until a defined limit is obtained.

For the case in point,
lim x->0 (e^-x-1)/(1-e^x)
= lim x->0 -e^-x/-e^x
= 1