Question Number 130223 by Study last updated on 23/Jan/21
$${li}\underset{{x}\rightarrow\overset{−} {\mathrm{5}}} {{m}lo}\underset{\frac{\mathrm{2}}{\mathrm{3}}} {{g}}\left(\mathrm{3}{x}−\mathrm{15}\right)=? \\ $$
Answered by EDWIN88 last updated on 23/Jan/21
$$\underset{{x}\rightarrow\mathrm{5}^{−} } {\mathrm{lim}}\:\mathrm{log}\:_{\mathrm{2}/\mathrm{3}} \left(\mathrm{3x}−\mathrm{15}\right)=\frac{\underset{{x}\rightarrow\mathrm{5}^{−} } {\mathrm{lim}ln}\:\left(\mathrm{3x}−\mathrm{15}\right)}{\mathrm{ln}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}=−\infty \\ $$
Commented by Study last updated on 23/Jan/21
$$\:{if}\:\:\:{x}<\mathrm{5}\:\:\:\:\:{log}\left(\mathrm{3}{x}−\mathrm{15}\right)={log}\left(−\right)=? \\ $$
Commented by EDWIN88 last updated on 23/Jan/21
