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Question Number 62200 by maxmathsup by imad last updated on 17/Jun/19

$$calculate{lim}_{x\to 0}\frac{ln(1+x+sinx)-ln(1+sin\left(2x\right))}{x^2}$$

Commented by maxmathsup by imad last updated on 18/Jun/19

$$letusehospitaltheoremletu\left(x\right)=ln(1+x+sinx)-ln(1+sin\left(2x\right)\Rightarrow$$$$u^{{}^{\prime }}\left(x\right)=\frac{1+cosx}{1+x+sinx}-\frac{2cosx}{1+sin\left(2x\right)}$$$$u^{{}^{″}}\left(x\right)=\frac{-sinx(1+x+sinx)-{(1+cosx)}^2}{{(1+x+sinx)}^2}-2\frac{-sinx(1+sin\left(2x\right))-cosx\left(2cosx\left(2x\right)\right)}{{(1+sin\left(2x\right))}^2}$$$$\Rightarrow {lim}_{x\to 0}{u}^{\left(2\right)}\left(x\right)=\frac{-4}{1}-2\frac{-2}{1}=-4+4=0$$$$v\left(x\right)=x^2\Rightarrow v^{{}^{\prime }}\left(x\right)=2x\Rightarrow v^{\left(2\right)}\left(x\right)=2\Rightarrow {lim}_{x\to 0}\frac{ln(1+x+sinx)-ln(1+sin\left(2x\right))}{x^2}$$$$={lim}_{x\to 0}\frac{u^{\left(2\right)}\left(x\right)}{v^{\left(2\right)}\left(x\right)}=0$$

Answered by tanmay last updated on 17/Jun/19

$$\underset{x\to 0}{\lim }\frac{ln\left(\frac{1+x+sinx}{1+sin2x}\right)}{x^2}$$$$\underset{x\to 0}{\lim }\frac{ln(1+\frac{1+x+sinx}{1+sin2x}-1)}{(\frac{1+x+sinx}{1+sin2x}-1)}\times \frac{\frac{x+sinx-sin2x}{1+sin2x}}{x^2}$$$$\underset{t\to 0}{\lim }\frac{ln(1+t)}{t}\times \underset{x\to 0}{\lim }\frac{1}{1+sin2x}\times \underset{x\to 0}{\lim }\frac{x+sinx-sin2x}{x^2}$$$$1\times 1\times \underset{x\to 0}{\lim }\frac{1+cosx-2cos2x}{2x}\left(\frac{0}{0}\right)$$$$=\underset{x\to 0}{\lim }\frac{-sinx+4sin2x}{2}=0$$$$$$

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