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Question Number 152431 by fotosy2k last updated on 28/Aug/21

Commented by fotosy2k last updated on 28/Aug/21

help pls

$${help}\:{pls} \\ $$

Commented by EDWIN88 last updated on 28/Aug/21

lim_(x→0^− ) f(x)=lim_(x→0^− ) (((e^(5x) −e^(2x) )/x)) =lim_(x→0^− ) (((e^(5x) −1)/x) +((1−e^(2x) )/x)) =lim_(x→0^− ) (((e^(5x) −1)/x))−lim_(x→0^− ) (((e^(2x) −1)/x)) = 5−2=3 lim_(x→0^+ ) f(x)=lim_(x→0^+ ) (3)=3 so lim_(x→0^− ) f(x)=lim_(x→0^+ ) f(x)=f(0)=3 it follows that f(x) continu at x=0

$$\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left(\frac{{e}^{\mathrm{5}{x}} −{e}^{\mathrm{2}{x}} }{{x}}\right) \\ $$$$\:=\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left(\frac{{e}^{\mathrm{5}{x}} −\mathrm{1}}{{x}}\:+\frac{\mathrm{1}−{e}^{\mathrm{2}{x}} }{{x}}\right) \\ $$$$\:=\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left(\frac{{e}^{\mathrm{5}{x}} −\mathrm{1}}{{x}}\right)−\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left(\frac{{e}^{\mathrm{2}{x}} −\mathrm{1}}{{x}}\right) \\ $$$$\:=\:\mathrm{5}−\mathrm{2}=\mathrm{3} \\ $$$$ \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\mathrm{3}\right)=\mathrm{3} \\ $$$${so}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{f}\left({x}\right)={f}\left(\mathrm{0}\right)=\mathrm{3} \\ $$$${it}\:{follows}\:{that}\:{f}\left({x}\right)\:{continu}\:{at}\:{x}=\mathrm{0} \\ $$

Commented by fotosy2k last updated on 30/Aug/21

thank you

$${thank}\:{you} \\ $$

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