Question Number 49817 by Aditya789 last updated on 11/Dec/18
$$\mathrm{prove}\:\mathrm{that}.\:\frac{\mathrm{a}^{\mathrm{r}} −\mathrm{1}}{\mathrm{r}}=\mathrm{1} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18
$${what}\:{is}\:{the}\:{relation}\:{between}\:{a}\:{and}\:{r}…{question} \\ $$$${not}\:{clear}… \\ $$
Answered by $@ty@m last updated on 11/Dec/18
$${Pl}.\:{check}\:{the}\:{question}. \\ $$$${In}\:{my}\:{opinion},\:{it}\:{should}\:{be}: \\ $$$${Prove}\:{that}\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{{r}} −\mathrm{1}}{{r}}=\mathrm{1} \\ $$$${Proof}: \\ $$$$\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{r}} −\mathrm{1}}{{r}}\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+{r}+\frac{{r}^{\mathrm{2}} }{\mathrm{2}!}+…−\mathrm{1}}{{r}}\: \\ $$$$=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{r}+\frac{{r}^{\mathrm{2}} }{\mathrm{2}!}+…}{{r}}=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{{r}}{\mathrm{2}!}+…\right)=\mathrm{1} \\ $$
Commented by $@ty@m last updated on 11/Dec/18
$${Similarly},\:{it}\:{can}\:{be}\:{shown}\:{that}: \\ $$$$\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}^{{r}} −\mathrm{1}}{{r}}=\mathrm{ln}\:{a} \\ $$