Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 49817 by Aditya789 last updated on 11/Dec/18

prove that. ((a^r −1)/r)=1

$$\mathrm{prove}\:\mathrm{that}.\:\frac{\mathrm{a}^{\mathrm{r}} −\mathrm{1}}{\mathrm{r}}=\mathrm{1} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

what is the relation between a and r...question  not clear...

$${what}\:{is}\:{the}\:{relation}\:{between}\:{a}\:{and}\:{r}...{question} \\ $$$${not}\:{clear}... \\ $$

Answered by $@ty@m last updated on 11/Dec/18

Pl. check the question.  In my opinion, it should be:  Prove that lim_(r→0)   ((e^r −1)/r)=1  Proof:  lim_(r→0) ((e^r −1)/r) = lim_(r→0) ((1+r+(r^2 /(2!))+...−1)/r)   =lim_(r→0) ((r+(r^2 /(2!))+...)/r)=lim_(r→0) (1+(r/(2!))+...)=1

$${Pl}.\:{check}\:{the}\:{question}. \\ $$$${In}\:{my}\:{opinion},\:{it}\:{should}\:{be}: \\ $$$${Prove}\:{that}\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{{r}} −\mathrm{1}}{{r}}=\mathrm{1} \\ $$$${Proof}: \\ $$$$\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{r}} −\mathrm{1}}{{r}}\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+{r}+\frac{{r}^{\mathrm{2}} }{\mathrm{2}!}+...−\mathrm{1}}{{r}}\: \\ $$$$=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{r}+\frac{{r}^{\mathrm{2}} }{\mathrm{2}!}+...}{{r}}=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{{r}}{\mathrm{2}!}+...\right)=\mathrm{1} \\ $$

Commented by $@ty@m last updated on 11/Dec/18

Similarly, it can be shown that:  lim_(r→0) ((a^r −1)/r)=ln a

$${Similarly},\:{it}\:{can}\:{be}\:{shown}\:{that}: \\ $$$$\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}^{{r}} −\mathrm{1}}{{r}}=\mathrm{ln}\:{a} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com