Pembahasan Limit UTUL UGM 2017 Matematika Ipa Kode 814

Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to -y} \frac{\tan x + \tan y}{\left(\frac{x^2-y^2}{-2y^2} \right) (1 - \tan x \tan y)} = .... $
A). $ -1 \, $ B). $ 1 \, $ C). $ 0 \, $ D). $ y \, $ E). $ -y $

$\spadesuit $ Konsep Dasar Limit fungsi Trigonometri
*). 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(\frac{x^2-y^2}{-2y^2} \right) (1 - \tan x \tan y)} \\ & = \displaystyle \lim_{x \to -y} \, \frac{-2y^2}{x^2-y^2} . \frac{\tan x + \tan y}{(1 - \tan x \tan y)} \\ & = \displaystyle \lim_{x \to -y} \, \frac{-2y^2}{x^2-y^2} . \tan (x + y) \\ & = \displaystyle \lim_{x \to -y} \, \frac{-2y^2}{(x-y)(x+y)} . \tan (x + y) \\ & = \displaystyle \lim_{x \to -y} \, \frac{\tan (x+y) }{(x+y)} . \frac{-2y^2}{(x-y) } \\ & = 1 . \frac{-2y^2}{(-y-y) } \\ & = \frac{-2y^2}{(-2y) } = y \end{align} $ .
Jadi, hasil limitnya adalah $ y . \, \heartsuit $

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