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 23584 by A1B1C1D1 last updated on 02/Nov/17

Answered by FilupES last updated on 02/Nov/17

lim_(x→0^± ) ((ln(3+x)−ln(x))/x)=(lim_(x→0^± ) (1/x))(lim_(x→0^± ) {ln(3+x)−ln(x)}) let M>0 (1) (1/x)>M for 0<x<(1/M) for large M, lim_(x→0^+ ) (1/x)=+∞ (2) −(1/x)<−M for −(1/M)<x<0 for large M, lim_(x→0^− ) (1/x)=−∞ lim_(x→0^± ) ((ln(3+x)−ln(x))/x)=(lim_(x→0^± ) (1/x))(lim_(x→0^± ) {ln(3+x)−ln(x)}) =(±∞)(lim_(x→0^± ) {ln(3+x)−ln(x)}) =(±∞)(ln(3)−ln(0)) =(±∞)(ln(3)−(−∞)) =(±∞)(∞) ∴lim_(x→0^± ) ((ln(3+x)−ln(x))/x)=±∞

$$\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{3}+{x}\right)−\mathrm{ln}\left({x}\right)}{{x}}=\left(\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\right)\left(\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\left\{\mathrm{ln}\left(\mathrm{3}+{x}\right)−\mathrm{ln}\left({x}\right)\right\}\right) \\ $$$$\: \\ $$$$\mathrm{let}\:{M}>\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{x}}>{M}\:\:\:\mathrm{for}\:\mathrm{0}<{x}<\frac{\mathrm{1}}{{M}} \\ $$$$\mathrm{for}\:\mathrm{large}\:{M},\:\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{1}}{{x}}=+\infty \\ $$$$\left(\mathrm{2}\right)\:−\frac{\mathrm{1}}{{x}}<−{M}\:\:\mathrm{for}\:\:−\frac{\mathrm{1}}{{M}}<{x}<\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{large}\:{M},\:\:\:\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\frac{\mathrm{1}}{{x}}=−\infty \\ $$$$\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{3}+{x}\right)−\mathrm{ln}\left({x}\right)}{{x}}=\left(\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\right)\left(\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\left\{\mathrm{ln}\left(\mathrm{3}+{x}\right)−\mathrm{ln}\left({x}\right)\right\}\right) \\ $$$$=\left(\pm\infty\right)\left(\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\left\{\mathrm{ln}\left(\mathrm{3}+{x}\right)−\mathrm{ln}\left({x}\right)\right\}\right) \\ $$$$=\left(\pm\infty\right)\left(\mathrm{ln}\left(\mathrm{3}\right)−\mathrm{ln}\left(\mathrm{0}\right)\right) \\ $$$$=\left(\pm\infty\right)\left(\mathrm{ln}\left(\mathrm{3}\right)−\left(−\infty\right)\right) \\ $$$$=\left(\pm\infty\right)\left(\infty\right) \\ $$$$\: \\ $$$$\therefore\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{3}+{x}\right)−\mathrm{ln}\left({x}\right)}{{x}}=\pm\infty \\ $$

Commented by A1B1C1D1 last updated on 02/Nov/17

Does this limit exist?

$$\mathrm{Does}\:\mathrm{this}\:\mathrm{limit}\:\mathrm{exist}? \\ $$

Commented by A1B1C1D1 last updated on 02/Nov/17

Side limits?

$$\mathrm{Side}\:\mathrm{limits}? \\ $$

Commented by ajfour last updated on 02/Nov/17

thanks, i suspected.

$${thanks},\:{i}\:{suspected}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com