lim-x-2-16-4-x-9-3-x-with-out-lophital-

Question Number 150202 by tabata last updated on 10/Aug/21

$${lim}_{{x}\rightarrow\mathrm{2}} \left(\frac{\mathrm{16}−\mathrm{4}^{{x}} }{\mathrm{9}−\mathrm{3}^{{x}} }\right)\:{with}\:{out}\:{lophital} \\ $$

Commented by tabata last updated on 10/Aug/21

$$?????? \\ $$

Answered by liberty last updated on 10/Aug/21

lim_(x→2) (((4^x (4^(2−x) −1))/(3^x (3^(2−x) −1))))=((4/3))^2 ×lim_(x→2) ((4^(2−x) −1)/(3^(2−x) −1)) =((16)/9)×lim_(x→2) ((e^((2−x)ln 4) −1)/(e^((2−x)ln 3) −1)) = ((16)/9)×((−ln 4.(4^(2−x) ))/(−ln 3(3^(2−x) ))) = ((16 ln 4)/(9 ln 3))

$$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left(\frac{\mathrm{4}^{\mathrm{x}} \left(\mathrm{4}^{\mathrm{2}−\mathrm{x}} −\mathrm{1}\right)}{\mathrm{3}^{\mathrm{x}} \left(\mathrm{3}^{\mathrm{2}−\mathrm{x}} −\mathrm{1}\right)}\right)=\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} ×\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{4}^{\mathrm{2}−\mathrm{x}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}−\mathrm{x}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{16}}{\mathrm{9}}×\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\left(\mathrm{2}−\mathrm{x}\right)\mathrm{ln}\:\mathrm{4}} −\mathrm{1}}{\mathrm{e}^{\left(\mathrm{2}−\mathrm{x}\right)\mathrm{ln}\:\mathrm{3}} −\mathrm{1}} \\ $$$$=\:\frac{\mathrm{16}}{\mathrm{9}}×\frac{−\mathrm{ln}\:\mathrm{4}.\left(\mathrm{4}^{\mathrm{2}−\mathrm{x}} \right)}{−\mathrm{ln}\:\mathrm{3}\left(\mathrm{3}^{\mathrm{2}−\mathrm{x}} \right)} \\ $$$$=\:\frac{\mathrm{16}\:\mathrm{ln}\:\mathrm{4}}{\mathrm{9}\:\mathrm{ln}\:\mathrm{3}} \\ $$

Commented by tabata last updated on 10/Aug/21

$${where}\:{the}\:{complete}\:{solution}\: \\ $$

Commented by tabata last updated on 10/Aug/21

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

Commented by tabata last updated on 10/Aug/21

$${step}\:\left(\mathrm{3}\right)\:{by}\:{lophital}\:? \\ $$

Answered by puissant last updated on 10/Aug/21

lim_(x→2) (((16−e^(xln4) )/(9−e^(xln3) ))) = lim_(x→2) ((−ln4e^(xln4) )/(−ln3e^(xln3) )) (hopital) ⇒ lim_(x→2) (((16−4^x )/(9−3^x )))= ((ln4)/(ln3))×(e^(2ln4) /e^(2ln3) ) = ((16)/9)log_3 (4)..

$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{2}} \left(\frac{\mathrm{16}−\mathrm{e}^{\mathrm{xln4}} }{\mathrm{9}−\mathrm{e}^{\mathrm{xln3}} }\right)\:=\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{2}} \frac{−\mathrm{ln4e}^{\mathrm{xln4}} }{−\mathrm{ln3e}^{\mathrm{xln3}} }\:\left(\mathrm{hopital}\right) \\ $$$$\Rightarrow\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{2}} \left(\frac{\mathrm{16}−\mathrm{4}^{\mathrm{x}} }{\mathrm{9}−\mathrm{3}^{\mathrm{x}} }\right)=\:\frac{\mathrm{ln4}}{\mathrm{ln3}}×\frac{\mathrm{e}^{\mathrm{2ln4}} }{\mathrm{e}^{\mathrm{2ln3}} } \\ $$$$=\:\frac{\mathrm{16}}{\mathrm{9}}\mathrm{log}_{\mathrm{3}} \left(\mathrm{4}\right).. \\ $$

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