Question #c91de

assuming you mean natural log here

#lim_{x to 0} (ln (1+x)) / x#

#=lim_{x to 0} (x - x^2/2 + mathcal(O)(x^3) ) / x# using the Taylor expansion for the numerator about x = 0

#=lim_{x to 0} 1 + mathcal(O)(x) = 1#

of you can hit it with L'Hopital as it's #0/0# indeterminate so

#lim_{x to 0} ((1/(1+x))) / 1 = 1/(1+x) = 1#

you can also look at it as

#lim_{x to 0} 1/xln (1+x)#

#lim_{x to 0} ln color(red)((1+x)^(1/x))#

and that bit in red is the definition of #e# - see Bernoulli's compound interest formula, well-covered on Wiki, so you have #ln e = 1#

you'd need a little bit more algebra to do that, ie lift the #ln# outside the limit, which you can do as it is continuous about the limit

We need to find
#lim_(x->0) log (1+x)/x#
Substituting #x=0# in the expression # log (1+x)/x#, we see that it is indeterminate as in the form #0/0 .#
Therefore apply L'Hospital's Rule. Differentiate the numerator and differentiate the denominator and then take the limit.

We obtain
#lim_(x->0) (1/ (1+x))/1#
#=>lim_(x->0) 1/ (1+x)#
#= 1/ (1+0)#
#= 1#