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
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I'm sorry I edit the post, there was a mistake in the question , – shir Jan 20 '21 at 19:01
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cot(X)= cos(x)/sin(X) – shir Jan 20 '21 at 19:01
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I'm sure your textbook (or a table) has enough many terms on the Taylor series of the tangent. $$ \lim_{x\to0}\frac{x-\tan x}{-17x^2\tan x}=?$$ First degree terms cancel out in the numerator. Then take out a common factor $x^3$. Done. – Jyrki Lahtonen Jan 20 '21 at 19:03
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Check https://math.stackexchange.com/questions/2186061/ out as well. – player3236 Jan 20 '21 at 19:05
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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 $\boxed{ \frac{1}{51}}$.
hamam_Abdallah
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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...$$x\to 0 \\\tan x \sim x+ \frac {x^3}{3}$$
SNEHIL SANYAL
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Khosrotash
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