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I need to find the limit: $$\lim_{x\to0}\frac{x\cot(x)-1}{-17x^2}$$

I've been tried L'Hopital and Taylor but it's not going so well.. thank in advance

shir
  • 81

2 Answers2

2

Hint

$$\frac{x}{\sin(x)}=\frac{x}{x-\frac{x^3}{6}(1+\epsilon(x))}$$

$$=\frac{1}{1-\frac{x^2}{6}(1+\epsilon(x))}$$

$$=1+\frac{x^2}{6}(1+\epsilon(x))$$

So,

$$x \;cot(x)=\frac{x}{\sin(x)}\cos(x)$$ $$=(1+\frac{x^2}{6})(1-\frac{x^2}{2})+x^2\epsilon(x)$$

$$=1-\frac{x^2}{2}+\frac{x^2}{6}+x^2\epsilon(x)$$

You will find $\boxed{ \frac{1}{51}}$.

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Welcome to MSE: $$\lim_{x\to0}\frac{x\cot(x)-1}{-17x^2}=\\ \lim_{x\to0}\frac{x\cot(x)-1}{-17x^2}\times \frac{\tan x}{\tan x}=\\ \lim_{x\to0}\frac{x-\tan x}{-17x^2\tan x}$$ then use...$$x\to 0 \\\tan x \sim x+ \frac {x^3}{3}$$

Khosrotash
  • 24,922