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Question Number 93200 by i jagooll last updated on 11/May/20

Commented by mathmax by abdo last updated on 11/May/20

$$letf\left(x\right)={(2^x+3^x-12)}^{tan\left(\frac{\pi x}{4}\right)}\Rightarrow ln\left(f\left(x\right)\right)=tan\left(\frac{\pi x}{4}\right)ln(2^x+3^x-12)$$$$changementt=x-2giveln\left(f\left(x\right)\right)=$$$$tan\left(\frac{\pi (t+2)}{4}\right)ln(2^{t+2}+3^{t+2}-12)=\frac{ln(42^t+9.3^t-12)}{tan\left(\frac{\pi t}{4}\right)}=g\left(t\right)$$$$letu\left(x\right)=ln(4.2^t+9.3^t-12)andv\left(t\right)=tan\left(\frac{\pi t}{4}\right)\left(hospital\right)$$$$wehave{lim}_{t\to 0}g\left(t\right)={lim}_{t\to 0}\frac{u^{{}^{\prime }}\left(x\right)}{v^{{}^{\prime }}\left(x\right)}$$$$u^{{}^{\prime }}\left(x\right)=\frac{{(4.2^t+9.3^t-12)}^{{}^{\prime }}}{4.2^t+9.3^t-12}=\frac{{4.e^{tln2}+9.e^{tln3}-12)}^{{}^{\prime }}}{4.2^t+9.3^t-12}$$$$=\frac{4ln\left(2\right)2^t+9ln\left(3\right).3^t}{4.2^t+93^t-12}\Rightarrow {lim}_{t\to 0}{u}^{{}^{\prime }}\left(x\right)=4ln\left(2\right)+9ln\left(3\right)$$$$v^{{}^{\prime }}\left(t\right)=\frac{\pi }{4}(1+{tan}^2\left(\frac{\pi t}{4}\right))\Rightarrow {lim}_{t\to 0}{v}^{{}^{\prime }}\left(t\right)=\frac{\pi }{4}\Rightarrow$$$${lim}_{t\to 0}g\left(t\right)=\frac{4}{\pi }(4ln\left(2\right)+9ln\left(3\right))={lim}_{x\to 2}lnf\left(x\right)\Rightarrow$$$${lim}_{x\to 2}f\left(x\right)=e^{\frac{4}{\pi }(4ln\left(2\right)+9ln\left(3\right))}$$

Commented by john santu last updated on 12/May/20

$$\mathrm{sorry}\mathrm{sir}.$$$$\frac{\ln (4.2^{\mathrm{t}}+9.3^{\mathrm{t}}-12)}{\cot \left(\frac{\pi (\mathrm{t}+2)}{4}\right)}\mathrm{should}\mathrm{be}$$$$\frac{\ln (4.2^{\mathrm{t}}+9.3^{\mathrm{t}}-12)}{-\tan \left(\frac{\pi \mathrm{t}}{4}\right)}$$

Commented by mathmax by abdo last updated on 12/May/20

$$yesforgetthesign-$$

Answered by john santu last updated on 12/May/20

$$\underset{\mathrm{w}\to 0}{\lim }\ln \mathrm{y}=\underset{\mathrm{w}\to 0}{\lim }\tan \left(\frac{\pi }{4}(\mathrm{w}+2)\right)\ln (2^{\mathrm{w}+2}+3^{\mathrm{w}+2}-12)$$$$=\underset{\mathrm{w}\to 0}{\lim }\frac{\ln (2^{\mathrm{w}+2}+3^{\mathrm{w}+2}-12)}{\cot (\frac{\pi }{2}+\frac{\pi \mathrm{w}}{4})}$$$$\underset{\mathrm{w}\to 0}{\lim }\frac{\ln (2^{\mathrm{w}+2}+3^{\mathrm{w}+2}-12)}{-\tan \left(\frac{\pi \mathrm{w}}{4}\right)}=$$$$\underset{\mathrm{w}\to 0}{\lim }\frac{2^{\mathrm{w}+2}\ln 2+3^{\mathrm{w}+2}\ln 3}{2^{\mathrm{w}+2}+3^{\mathrm{w}+2}-12}\times (-\frac{4}{\pi }\cos {}^2\left(\frac{\pi \mathrm{w}}{4}\right))$$$$\underset{\mathrm{w}\to 0}{\lim }\ln \mathrm{y}=-\frac{4}{\pi }(4\ln 2+9\ln 3)$$$$\mathrm{y}={\mathrm{e}}^{-\frac{4}{\pi }(4\ln 2+9\ln 3)}.$$$$$$

Commented by i jagooll last updated on 12/May/20

$$\mathrm{yeahh}...$$

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