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Question Number 82494 by niroj last updated on 21/Feb/20

$$$$$$\mathbf{Find}\mathbf{the}\mathbf{limit}$$$$\underset{\mathbf{x}\to 0}{\stackrel{\mathbf{lim}}{}}{\left(\frac{\mathbf{tan}\mathbf{x}}{\mathbf{x}}\right)}^{\frac{1}{{\mathbf{x}}^2}}$$

Commented by abdomathmax last updated on 21/Feb/20

$$letf\left(x\right)={\left(\frac{tanx}{x}\right)}^{\frac{1}{x^2}}\Rightarrow f\left(x\right)=e^{\frac{1}{x^2}ln\left(\frac{tanx}{x}\right)}$$$$u\left(x\right)=tanx=u\left(0\right)+\frac{x}{1!}{u}^{\left(1\right)}\left(o\right)+\frac{x^2}{2!}{u}^{\left(2\right)}\left(0\right)+\frac{x^3}{3!}{u}^{\left(3\right)}\left(0\right)+...$$$$u^{{}^{\prime }}\left(x\right)=1+{tan}^{2}x\Rightarrow u^{{}^{\prime }}\left(0\right)=1$$$$u^{\left(2\right)}\left(x\right)=2tanx(1+{tan}^{2}x)\Rightarrow u^{\left(2\right)}\left(0\right)=0$$$$u^{\left(3\right)}\left(x\right)=2(1+{tan}^{2}x)(1+{tan}^{2}x)+$$$$+\left(2tanx\right)\left(2tanx(1+{tan}^{2}x)\right)\Rightarrow u^{\left(3\right)}\left(0\right)=2\Rightarrow$$$$tanx=x+\frac{x^3}{3}+o\left(x^5\right)\Rightarrow \frac{tanx}{x}=1+\frac{x^2}{3}+o\left(x^4\right)\Rightarrow$$$$ln\left(\frac{tanx}{x}\right)=ln(1+\frac{x^2}{3}+o\left(x^4\right))\sim \frac{x^2}{3}\Rightarrow$$$$\frac{1}{x^2}ln\left(\frac{tanx}{x}\right)\sim \frac{1}{3}\Rightarrow {lim}_{x\to 0}f\left(x\right)=e^{\frac{1}{3}}{=}^3\sqrt{e}$$

Commented by jagoll last updated on 22/Feb/20

$$\underset{x\to 0}{\lim }{(1+\frac{\tan x}{x}-1)}^{\frac{1}{x^2}}=$$$$\underset{x\to 0}{\lim }{(1+\frac{\tan x-x}{x})}^{\frac{1}{x^2}}=$$$$e{}^{\underset{x\to 0}{\lim }\left(\frac{\tan x-x}{x^3}\right)}=e^{\underset{x\to 0}{\lim }\left(\frac{\sec {}^{2}x-1}{3x^2}\right)}$$$$=e^{\underset{x\to 0}{\lim }\left(\frac{1-\cos {}^{2}x}{3x^2\cos {}^{2}x}\right)}=e^{\frac{1}{3}}$$

Commented by john santu last updated on 22/Feb/20

$$nice$$

Commented by niroj last updated on 22/Feb/20

$$great.$$

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