How do you find the limit of #(e^x+x)^(1/x)# as x approaches 0 using l'hospital's rule?

Question

34140 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-23T11:34:21

Rewrite it as #e^(ln(e^x+x)^(1/x)# and find #lim_(xrarr0)ln(e^x+x)^(1/x)#.

#lim_(xrarr0)ln(e^x+x)^(1/x) = lim_(xrarr0)1/xln(e^x+x)#

# = lim_(xrarr0)ln(e^x+x)/x#.

This limit has indeterminate form #0/0#. #" "# (#ln(e^0+0) = ln1 =0#)

Apply l"Hospital's rule:

Find
#lim_(xrarr0) (1/(e^x+x)(e^x+1))/1 = lim_(xrarr0) (e^x+1)/(e^x+x) =2#.

As #xrarr0#, the exponent #ln(e^x+x)^(1/x)rarr2#, so

#lim_(xrarr0)(e^x+x)^(1/x) = lim_(xrarr0)e^(ln(e^x+x)^(1/x) = e^2#