# proof-lim-h-0-x-h-1-h-ln-x-

Question Number 11549 by Nayon last updated on 28/Mar/17
$$\:\:\:\: \\$$$${proof} \\$$$${lim} \\$$$${h}\rightarrow\mathrm{0}^{\frac{{x}^{{h}} −\mathrm{1}}{{h}}={ln}\left({x}\right)} \\$$
Answered by mrW1 last updated on 28/Mar/17
$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{{h}} −\mathrm{1}}{{h}}=\frac{\mathrm{0}}{\mathrm{0}} \\$$$$\Rightarrow\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{{h}} −\mathrm{1}}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{d}\left({x}^{{h}} −\mathrm{1}\right)}{{dh}}}{\frac{{d}\left({h}\right)}{{dh}}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{d}\left({x}^{{h}} \right)}{{dh}} \\$$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{d}\left({e}^{{h}\mathrm{ln}\:{x}} \right)}{{dh}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{{h}\mathrm{ln}\:{x}} ×\mathrm{ln}\:{x}=\mathrm{ln}\:{x} \\$$
Commented by Nayon last updated on 28/Mar/17
$${i}\:{could}\:{not}\:{understAnd}\:{the}\:\mathrm{2}{nd}\:{line} \\$$
Commented by mrW1 last updated on 28/Mar/17
$${l}'{hopital}'{s}\:{rule}\:{says}: \\$$$${if}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{f}\left({x}\right)=\mathrm{0}\:\left({or}\:\infty\right)\:{and}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{g}\left({x}\right)=\mathrm{0}\:\left({or}\:\infty\right) \\$$$${then} \\$$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}=\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{f}'\left({x}\right)}{{g}'\left({x}\right)} \\$$