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Question Number 80343 by jagoll last updated on 02/Feb/20

$$\underset{x\to 0}{\lim }\frac{ln(\tan x+1)-\sin x}{x\sin x}$$

Commented by abdomathmax last updated on 02/Feb/20

$$wehave{ln}^{{}^{\prime }}(1+u)=\frac{1}{1+u}=1-u+o\left(u^2\right)(u\sim 0)$$$$\Rightarrow ln(1+u)=u-\frac{u^2}{2}+o\left(u^3\right)$$$$sinx=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}\Rightarrow sinx=x-\frac{x^3}{3!}+o\left(x^3\right)$$$$ln(1+tanx)\sim ln(1+x)=x-\frac{x^2}{2}+o\left(x^3\right)\Rightarrow$$$$ln(1+tanx)-sinx\sim x-\frac{x^2}{2}-x+\frac{x^3}{3!}=\frac{x^3}{6}-\frac{x^2}{2}$$$$xsinx\sim x^2\Rightarrow \frac{ln(1+tanx)-sinx}{xsinx}\sim \frac{x}{6}-\frac{1}{2}\Rightarrow$$$${lim}_{x\to 0}\frac{ln(1+tanx)-sinx}{xsinx}=-\frac{1}{2}$$

Commented by john santu last updated on 02/Feb/20

$$L^{\prime }Hopital$$$$\underset{x\to 0}{\lim }\frac{\frac{\sec {}^{2}x}{1+\tan x}-\cos x}{x\cos x+\sin x}=$$$$A=\underset{x\to 0}{\lim }\frac{1}{1+\tan x}=1$$$$B=\underset{x\to 0}{\lim }\frac{\sec {}^{2}x-\cos x(1+\tan x)}{x\cos x+\sin x}$$$$B=\underset{x\to 0}{\lim }\frac{\sec {}^{2}x-\cos x-\sin x}{x\cos x+\sin x}$$$$B=\underset{x\to 0}{\lim }\frac{1}{\cos {}^{2}x}\times \underset{x\to 0}{\lim }\frac{1-\cos {}^{3}x-(1-\sin {}^{2}x)\sin x}{x\cos x+\sin x}$$$$B=\underset{x\to 0}{\lim }\frac{3\cos {}^{2}x\sin x-\cos x+3\sin {}^{2}x\cos x}{-x\sin x+\cos x+\cos x}$$$$B=-\frac{1}{2}\iff A\times B=-\frac{1}{2}$$$$$$

Commented by jagoll last updated on 02/Feb/20

$$thankyoumister$$

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