How do you find the limit of #[(1/ln(x+1)) - (1/x)]# as x approaches 0?

Question

8518 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-07T09:04:33

#1/2#

Making the substitution #e^y equiv x + 1# we have

#lim_{x->0}(1/log_e(x+1)-1/x) equiv lim_{y->0}(1/log_e(e^y)-1/(e^y-1)) #

but

#1/log_e(e^y)-1/(e^y-1)=1/y-1/(e^y -1) = (e^y-1-y)/(y(e^y-1)#

We know that

#e^y = 1 + y + y^2/(2!)+y^3/(3!)+cdots#

so

#(e^y-1-y)/(y(e^y-1)) = ( y^2/(2!)+y^3/(3!)+...)/(y^2+y^3/(2!)+y^4/(3!)+cdots) = (1/2+y/(3!)+cdots)/(1+y/(2!)+y^2/(3!)+ cdots)#

so

#lim_{y->0}(e^y-1-y)/(y(e^y-1)) equiv lim_{x->0}(1/log_e(x+1)-1/x) = 1/2#