Note that lim_(urarroo)(1+1/u)^u = e.
If lim_(xrarroo)f(x) = 0, then
lim_(xrarroo)(1+f(x)) = lim_(xrarroo)[(1+1/(1/f(x)))^(1/f(x))]^f(x)
By the limit stated above, with u = 1/f(x), we have
lim_(xrarroo)u = oo, and so,
lim_(xrarroo)[(1+1/(1/f(x)))^(1/f(x))] = lim_(urarroo)(1+1/u)^u = e
So
lim_(xrarroo)(1+f(x))^g(x) = lim_(xrarroo)([(1+1/(1/f(x)))^(1/f(x))]^f(x))^g(x)
= lim_(xrarroo) ([e]^f(x))^g(x)
= lim_(xrarroo) e^(f(x)g(x))
= e^(lim_(xrarroo)f(x)g(x))
Notes
lim_(urarroo)(1+1/u)^u = e is sometimes taken as the definition of e.
It can be verified using l'Hospital's Rule
ln((1+1/u)^u) = u ln(1+1/u) = ln(1+1/u)/(1/u) which takes indeterminate form (-oo)/oo as xrarroo
L'Hospitals's rule asks us to eveluate
lim_(urarroo)(1/(1+1/u)(-1/u^2))/(-1/u^2) which evaluates to 1.
Since the limit of the ln is 1, the limit of the expression is e.
Another Limit Result
If lim_(xrarra)f(x)=1 and lim_(xrarra)g(x) = oo,
then we can evaluate lim_(xrarra)(f(x))^(g(x)) using lim_(urarroo)(1+1/u)^u = e to get
lim_(xrarra)(f(x))^(g(x)) = e^(lim_(xrarra)(f(x)-1)g(x)).
Outline of proof:
As xrarra,
we have f(x) rarr 1, so
(f(x)-1)rarr0^+, and
1/(f(x)-1) rarroo
Therefore, as xrarra, (with u = 1/(f(x)-1)) we have
(1+1/(1/(f(x)-1)))^(1/(f(x)-1)) rarr e
We can conclude that
lim_(xrarra)(f(x))^(g(x)) = lim_(xrarra)e^((f(x)-1)g(x))
= e^(lim_(xrarra)(f(x)-1)g(x))
Although this equality holds for any g(x), it is only interesting if lim_(xrarr)g(x) = oo " or " -oo. Otherwise we get a triviality 1^g(x) = e^0
