What is the limit of #(e^x + x)^(1/x)# as x approaches infinity?
2 Answers
Explanation:
Explanation:
#L=lim_(xrarroo)(e^x+x)^(1/x)#
A helpful tool in finding limits with exponential functions like this is to use the equivalence
#L=lim_(xrarroo)e^(ln(e^x+x)^(1/x))#
This can be rewritten using the log rule
#L=lim_(xrarroo)e^(1/xln(e^x+x))#
We can move the limit inside of the exponential function:
#L=e^(lim_(xrarroo)(ln(e^x+x)/x)#
Notice that
#L=e^(lim_(xrarroo)((d/dxln(e^x+x))/(d/dxx))#
#L=e^(lim_(xrarroo)((e^x+1)/(e^x+x))/1#
#L=e^(lim_(xrarroo)(e^x+1)/(e^x+x)#
You may immediately recognize that this limit is
#L=e^(lim_(xrarroo)(e^x/(e^x+1))#
And L'Hopital's once more, as we're still in
#L=e^(lim_(xrarroo)(e^x/e^x)#
#L=e^(lim_(xrarroo)1)#
#L=e^1#
#L=e#