What is the limite x approaches 0 ((x/(x+1))^x?

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Answer 1 · 2018-03-05T16:33:40

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

Write the function as:

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

and consider the limit:

#lim_(x->0) xln(x/(x+1))#

transform it as:

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

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

#lim_(x->0) xln(x/(x+1)) = - lim_(x->0) ln(1+1/x)/(1/x) #

which is in the form #oo/oo# so we can apply l'Hospital's rule:

#lim_(x->0) xln(x/(x+1)) = - lim_(x->0) (d/dx ln(1+1/x))/(d/dx (1/x) )#

#lim_(x->0) xln(x/(x+1)) = - lim_(x->0) 1/(1+1/x) (-1/x^2)/(-1/x^2 )#

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

Now as #e^x# is continuous for #x=0#:

#lim_(x->0) (x/(x+1))^x = e^ (lim_(x->0)(xln(x/(x+1)))) =e^0=1#

graph{(x/(x+1))^x [-10, 10, -.1, 1]}