Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

lim_(x→2) (((16−4^x )/(9−3^x ))) with out lophital

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

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

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

Commented by tabata last updated on 10/Aug/21

thank you alot

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

Commented by tabata last updated on 10/Aug/21

step (3) by lophital ?

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

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)..

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com