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#
