Question-23584

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