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Question-145010




Question Number 145010 by mim24 last updated on 01/Jul/21
Answered by ArielVyny last updated on 01/Jul/21
consider f(x)=a^x      f(0)=1  then f′(x)=ln(a)a^x   lim_(x→0) ((a^x −1)/x)=lim((f(x)−f(0))/(x−0))=f′(0)=lna
$${consider}\:{f}\left({x}\right)={a}^{{x}} \:\:\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${then}\:{f}'\left({x}\right)={ln}\left({a}\right){a}^{{x}} \\ $$$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{a}^{{x}} −\mathrm{1}}{{x}}={lim}\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}−\mathrm{0}}={f}'\left(\mathrm{0}\right)={lna} \\ $$
Answered by mathmax by abdo last updated on 01/Jul/21
f(x)=((a^x −1)/x) ⇒f(x)=((e^(xloga) −1)/x)  we have  e^(xloga)  ∼1+xloga ⇒  a^x −1∼xloga ⇒((a^x −1)/x)∼loga ⇒lim_(x→0) f(x)=loga
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}^{\mathrm{x}} −\mathrm{1}}{\mathrm{x}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{e}^{\mathrm{xloga}} −\mathrm{1}}{\mathrm{x}}\:\:\mathrm{we}\:\mathrm{have}\:\:\mathrm{e}^{\mathrm{xloga}} \:\sim\mathrm{1}+\mathrm{xloga}\:\Rightarrow \\ $$$$\mathrm{a}^{\mathrm{x}} −\mathrm{1}\sim\mathrm{xloga}\:\Rightarrow\frac{\mathrm{a}^{\mathrm{x}} −\mathrm{1}}{\mathrm{x}}\sim\mathrm{loga}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{loga} \\ $$

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