Question

Valoarea limitei cand x tinde la 0 din:
$( \frac{1}{ {x}^{2} } - {ctg}^{2} x)$
dau coroana, multumesc!
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Answer 1

$\lim\limits_{x\to 0}\Big(\dfrac{1}{x^2}-\text{ctg}^2 x\Big) = \lim\limits_{x\to 0}\left(\dfrac{1}{x^2}-\dfrac{1}{\tan^2 x}\right) = \\ \\ = \lim\limits_{x\to 0}\left(\dfrac{\tan^2 x-x^2}{x^2\tan^2 x}\right) = \lim\limits_{x\to 0}\left(\dfrac{\tan^2 x-x^2}{x^2\tan^2 x}\cdot \dfrac{x^2}{x^2}\right) =$

$=\lim\limits_{x\to 0}\left(\dfrac{\tan^2 x-x^2}{x^4}\right) =\lim\limits_{x\to 0}\dfrac{(\tan x+x)(\tan x-x)}{x\cdot x^3} = \\ \\= \lim\limits_{x\to 0}\dfrac{\tan x + x}{x}\cdot \lim\limits_{x\to 0}\dfrac{\tan x - x}{x^3} = 2\cdot \lim\limits_{x\to 0}\dfrac{\tan x - x}{x^3} = \\ \\ = 2\cdot \lim\limits_{x\to 0}\dfrac{\dfrac{1}{\cos ^2 x}-1}{3x^2} = 2\cdot \lim\limits_{x\to 0}\dfrac{-2\cos^{-3}x\cdot (-\sin x)}{6x} =$

$= 2\cdot \lim\limits_{x\to 0}\left(\dfrac{2\cos^{-3}x}{6}\cdot \dfrac{\sin x}{x}\right) = 2\cdot \dfrac{2}{6} = \boxed{\dfrac{2}{3}}\\ \\ \\ \lim\limits_{x\to 0}\dfrac{\tan x}{x} = 1,\quad \lim\limits_{x\to 0}\dfrac{\sin x}{x} = 1$