What is the limit of # (1+2x)^(1/x)# as x approaches infinity?

1 Answer
Feb 26, 2017

1

Explanation:

#lim_(x to oo) (1+2x)^(1/x)#

We can use exponentials and logs here:

#= lim_(x to oo) e^( ln(1+2x)^(1/x))#

#= lim_(x to oo) e^( ( ln(1+2x))/x)#

Because #e# is a continuous function in #(-oo, oo)#, we can also say this:

#= e^( lim_(x to oo) color(red)(( ln(1+2x))/x))#

....and we can focus on the term in red.

We know straightaway that, because the log will grow more slowly than the #x# term, then the term in red will go to zero; and so the limit is #1#.

We can use L'Hopital's Rule to see this out as this is in #oo/oo# indeterminate form. One application of L'H takes us here:

# ln(1+2x)/x to (2/(1+2x))/1 = 2/(1+ 2x)#

And so our limit is:

#= e^( lim_(x to oo) ( 2/(1+2x)))#

#= e^( lim_(x to oo) 0) = 1#