Question

A particular issue with the derivative of ln(x) using the definition of derivative...

Answer 1

\[\lim_{h \rightarrow 0}\frac{ \ln(x+h)-\ln(x) }{ h }=\lim_{h \rightarrow 0}\ \ln[(1+\frac{h }{x})^{1/h}]\]

now when you do a substitution like h=x/v then you have:

\[\lim_{x/v \rightarrow 0}\ \ln[(1+\frac{1 }{v})^{v/x}]=\lim_{v/x \rightarrow \infty}\ \frac{ 1 }{ x } \ln[(1+\frac{1 }{v})^{v}]\]

And I can see how that _should_ just leave (1/x)(ln(e)) but I can't really be sure that the v/x--> infinity part can really be assumed to just be the same as v-->infinity. Seems like the x plays a role here in taking the limit so it's sort of weird that it can come out in the end if you make this substitution.

Answer 2

\[\frac{ d }{ dx }(lnx)=\lim_{h \rightarrow 0}\frac{ \ln(x+h)-lnx }{ h }=\lim_{h \rightarrow 0}\frac{ \ln(\frac{ x+h }{ x } )}{ h }\]
\[=\lim_{h \rightarrow 0}\frac{ \ln(1+\frac{ h }{ x } )}{ h }\]
\[=\lim_{h \rightarrow 0}\frac{ \ h \ln(1+\frac{ h }{ x })^{\frac{ x }{ h}}}{ xh }=\lim_{h \rightarrow 0}\frac{ \ \ln(1+\frac{ h }{ x })^{\frac{ x }{ h}}}{ x }\]=\[=\frac{ 1 }{ x} \lim_{h \rightarrow 0}\ln(1+\frac{ h }{ x })^{\frac{ x }{ h}}=\frac{ 1 }{ x }\ln(\lim_{h \rightarrow 0}(1+\frac{ h }{ x})^{\frac{ x }{ h }})\]
u=h/x;

\[\frac{ 1 }{ x }\ln(\lim_{u \rightarrow 0}(u+1)^{\frac{ 1 }{ u}})=\frac{ 1 }{ x}\ln(e)=\frac{ 1 }{ x }\]

Answer 3

Thanks, I think that answers it. I just needed to multiply by x/x and bring the x inside while leaving the 1/x outside, then make the substitution basically, so there are no problems with the limit.