What is the limit as x approaches infinity of #e^x#?

Question

William Abdalla · Accepted Answer

Another perspective...

#color(white)()# As a Real function Treating #e^x# as a function of Real values of #x#, it has the following properties: The domain of #e^x# is the whole of #RR#. The range of #e^x# is #(0, oo)#. #e^x# is continuous on the whole of #RR# and infinitely differentiable, with #d/(dx) e^x = e^x#. #e^x# is one to one, so has a well defined inverse function (#ln x#) from #(0, oo)# onto #RR#. #lim_(x->+oo) e^x = +oo# #lim_(x->-oo) e^x = 0# At first sight this answers the question, but what about Complex values of #x#? #color(white)()# As a Complex function Treated as a function of Complex values of #x#, #e^x# has the properties: The domain of #e^x# is the whole of #CC#. The range of #e^x# is #CC "\" { 0 }#. #e^x# is continuous on the whole of #CC# and infinitely differentiable, with #d/(dx) e^x = e^x#. #e^x# is many to one, so has no inverse function. The definition of #ln x# can be extended to a function from #CC "\" { 0 }# into #CC#, typically onto #{ x + iy : x in RR, y in (- pi, pi] }#. What do we mean by the limit of #e^x# as #x -> "infinity"# in this context? From the origin, we can head off towards "infinity" in all sorts of ways. For example, if we just set off along the imaginary axis, the value of #e^x# just goes round and around the unit circle. If we choose any complex number #c = r(cos theta + i sin theta)#, then following the line #ln r + it# for #t in RR# as #t->+oo#, the value of #e^(ln r + it)# will take the value #c# infinitely many times. We can project the Complex plane onto a sphere called the Riemann sphere #CC_oo#, with an additional point called #oo#. This allows us to picture the "neighbourhood of #oo#" and think about the behaviour of the function #e^x# there. From our preceding observations, #e^x# takes every non-zero complex value infinitely many times in any arbitrarily small neighbourhood of #oo#. That is called an essential singularity at infinity.

Scarlett Allison · Answer

Exaplanation using logarithms.

The limit does not exist because as #x# increases without bond, #e^x# also increases without bound. #lim_(xrarroo)e^x = oo#. Te xplanation of why will depand a great deal on the definitions of #e^x# and #lnx# with which you are working. I like to define #lnx = int_1^x 1/t dt# for #x > 0#, then prove that #lnx# is invertible (has an inverse) and define #e^x# as the inverse of #lnx#. Since #d/dx(lnx) = 1/x#, it is clear that #lnx# is increasing# In the process, I also prove to my students that that #lim_(xrarr00)lnx = oo#. (The proof uses ideas similar to those used in showing that the harmonic series diverges.) Knowing these things allows us to reason as follows. For any positive Real number #T#, #lnT# is defined and for all #x >= lnT#, we have #e^x >= e^(lnT) = T#. That is, for every #T#, there is an #M# (namely #lnT# such that if #x >= M, then #f(x) >= T# Therefore, by definition, #lim_(xrarroo)e^x = oo#

Christopher Allers · Answer

undefined The limit as x approaches infinity of e^x is infinity.