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Question-184960




Question Number 184960 by mathlove last updated on 14/Jan/23
Answered by cortano1 last updated on 15/Jan/23
L=−(9/4)(ln 4−1)
$${L}=−\frac{\mathrm{9}}{\mathrm{4}}\left(\mathrm{ln}\:\mathrm{4}−\mathrm{1}\right) \\ $$
Commented by mathlove last updated on 15/Jan/23
solution???
$${solution}??? \\ $$
Answered by cortano1 last updated on 15/Jan/23
 lim_(x→1)  ((ln (x+1)−ln 2)/( ((7+x))^(1/3) −(√(1+3x)))) × lim_(x→1 ) ((3(4^(x−1) −x))/(sin (x−1)))  L_1 =lim_(t→0) ((ln (t+2)−ln 2)/( ((t+8))^(1/3) −(√(4+3t))))   L_1 = lim_(t→0) ((1/(t+2))/((1/(3 (((t+8)^2 ))^(1/3) ))−(3/(2(√(4+3t))))))   L_1 = ((1/2)/((1/(12))−(3/4))) =(6/(1−9)) =−(3/4)  L_2 =lim_(x→1)  ((3(4^(x−1) −x))/(sin (x−1)))= 3.lim_(x→1)  ((4^(x−1) −1−(x−1))/(x−1))   L_2 = 3(ln 4−1)  ∴ L=L_1 ×L_2 =−(9/4)(ln 4−1)
$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left({x}+\mathrm{1}\right)−\mathrm{ln}\:\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\mathrm{7}+{x}}−\sqrt{\mathrm{1}+\mathrm{3}{x}}}\:×\:\underset{{x}\rightarrow\mathrm{1}\:} {\mathrm{lim}}\frac{\mathrm{3}\left(\mathrm{4}^{{x}−\mathrm{1}} −{x}\right)}{\mathrm{sin}\:\left({x}−\mathrm{1}\right)} \\ $$$${L}_{\mathrm{1}} =\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left({t}+\mathrm{2}\right)−\mathrm{ln}\:\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{{t}+\mathrm{8}}−\sqrt{\mathrm{4}+\mathrm{3}{t}}}\: \\ $$$${L}_{\mathrm{1}} =\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{{t}+\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{\left({t}+\mathrm{8}\right)^{\mathrm{2}} }}−\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{4}+\mathrm{3}{t}}}}\: \\ $$$${L}_{\mathrm{1}} =\:\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{3}}{\mathrm{4}}}\:=\frac{\mathrm{6}}{\mathrm{1}−\mathrm{9}}\:=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${L}_{\mathrm{2}} =\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{4}^{{x}−\mathrm{1}} −{x}\right)}{\mathrm{sin}\:\left({x}−\mathrm{1}\right)}=\:\mathrm{3}.\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{4}^{{x}−\mathrm{1}} −\mathrm{1}−\left({x}−\mathrm{1}\right)}{{x}−\mathrm{1}} \\ $$$$\:{L}_{\mathrm{2}} =\:\mathrm{3}\left(\mathrm{ln}\:\mathrm{4}−\mathrm{1}\right) \\ $$$$\therefore\:{L}={L}_{\mathrm{1}} ×{L}_{\mathrm{2}} =−\frac{\mathrm{9}}{\mathrm{4}}\left(\mathrm{ln}\:\mathrm{4}−\mathrm{1}\right) \\ $$
Commented by mathlove last updated on 15/Jan/23
thanks dear
$${thanks}\:{dear} \\ $$

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