Question Number 152431 by fotosy2k last updated on 28/Aug/21
Commented by fotosy2k last updated on 28/Aug/21
$${help}\:{pls} \\ $$
Commented by EDWIN88 last updated on 28/Aug/21
$$\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left(\frac{{e}^{\mathrm{5}{x}} −{e}^{\mathrm{2}{x}} }{{x}}\right) \\ $$$$\:=\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left(\frac{{e}^{\mathrm{5}{x}} −\mathrm{1}}{{x}}\:+\frac{\mathrm{1}−{e}^{\mathrm{2}{x}} }{{x}}\right) \\ $$$$\:=\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left(\frac{{e}^{\mathrm{5}{x}} −\mathrm{1}}{{x}}\right)−\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left(\frac{{e}^{\mathrm{2}{x}} −\mathrm{1}}{{x}}\right) \\ $$$$\:=\:\mathrm{5}−\mathrm{2}=\mathrm{3} \\ $$$$ \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\mathrm{3}\right)=\mathrm{3} \\ $$$${so}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{f}\left({x}\right)={f}\left(\mathrm{0}\right)=\mathrm{3} \\ $$$${it}\:{follows}\:{that}\:{f}\left({x}\right)\:{continu}\:{at}\:{x}=\mathrm{0} \\ $$
Commented by fotosy2k last updated on 30/Aug/21
$${thank}\:{you} \\ $$