Find the limit lim x→0^+ (1+x^2)^1/x?

find the lim#lim x→0^+ (1+x^2)^(1/x)#

1 Answer
Apr 12, 2018

#1#

Explanation:

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

plugging #x=0# directly makes #0/0#, so use l'hopital's rule

#ln(y)=lim_(xrarr0^+)(d/dx(ln(1+x^2)))/(d/dx(x))#

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

plugging #x=0# directly makes #0/1=0#

#ln(y)=0#
#y=e^0=1=lim_(xrarr0^+)(1+x^2)^(1/x)#

you can see #y# approaches 1 as #x# approaches 0
graph{(1+x^2)^(1/x) [-7.024, 7.024, -3.51, 3.513]}