Pembahasan Limit Trigonometri UM UGM 2004 Matematika Ipa

Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to y } \frac{\tan x - \tan y}{\left(1 - \frac{x}{y}\right)(1 + \tan x. \tan y) } = .... $
A). $ -1 \, $ B). $ 1 \, $ C). $ 0 \, $ D). $ y \, $ E). $ -y \, $

$\spadesuit $ Konsep Dasar
*). Sifat limit trigonometri :
$ \displaystyle \lim_{x \to k } \frac{\tan f(x)}{f(x)} = 1 $
dengan syarat $ f(k) = 0 $
*). RUmus trigonometri :
$ \tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x. \tan y} $

$\clubsuit $ Pembahasan
*). Menyelesaikan soal :
$ \begin{align} & \displaystyle \lim_{x \to y } \frac{\tan x - \tan y}{\left(1 - \frac{x}{y}\right)(1 + \tan x. \tan y) } \\ & = \displaystyle \lim_{x \to y } \frac{\tan x - \tan y}{\left(\frac{y}{y} - \frac{x}{y}\right)(1 + \tan x. \tan y) } \\ & = \displaystyle \lim_{x \to y } \frac{\tan x - \tan y}{\left(\frac{y - x }{y} \right)(1 + \tan x. \tan y) } \\ & = \displaystyle \lim_{x \to y } \frac{1}{\left(\frac{y - x }{y} \right)} . \frac{\tan x - \tan y}{(1 + \tan x. \tan y) } \\ & = \displaystyle \lim_{x \to y } \frac{y}{y-x} . \tan (x - y) \\ & = \displaystyle \lim_{x \to y } \frac{y}{-(x-y)} . \tan (x - y) \\ & = \displaystyle \lim_{x \to y } -y . \frac{\tan (x - y)}{(x-y)} \\ & = -y . 1 = - y \end{align} $
Jadi, hasil limitnya adalah $ -y . \, \heartsuit $

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