NakafaBelajar gratis dan berkualitas.

Nakafa School

Mata pelajaran

- Sarjana

Latihan

- TKA
- SNBT

Suci

Al Quran

Artikel

Politik

Komunitas
Tentang

Command Palette

Search for a command to run...

Soal 13

Nomor 13

Nilai dari $\lim_{x \to -y} \frac{\tan x + \tan y}{\frac{x^2 - y^2}{-2y^2} \cdot (1 - \tan x \tan y)}$ adalah....

$-1$

$1$

$0$

$y$

$-y$

Pembahasan

Ingat konsepnya bahwa

\lim_{x \to a} \frac{\tan m(x - a)}{n(x - a)} = \frac{m}{n}

Maka

\lim_{x \to -y} \frac{\tan x + \tan y}{\frac{x^2 - y^2}{-2y^2} \cdot (1 - \tan x \tan y)}

= \lim_{x \to -y} \frac{1}{\frac{x^2 - y^2}{-2y^2}} \cdot \frac{\tan x + \tan y}{1 - \tan x \tan y}

= \lim_{x \to -y} \frac{-2y^2}{x^2 - y^2} \cdot \tan(x + y)

= \lim_{x \to -y} \frac{-2y^2}{(x - y)(x + y)} \cdot \tan(x + y)

= \lim_{x \to -y} \frac{-2y^2}{x - y} \cdot \frac{\tan(x + y)}{x + y}

Karena $x \to -y$ maka $x + y \to 0$ , sehingga $\lim_{x + y \to 0} \frac{\tan(x + y)}{x + y} = 1$ . Maka

= \lim_{x \to -y} \frac{-2y^2}{x - y}

= \frac{-2y^2}{-y - y}

= \frac{-2y^2}{-2y}

= y

Komentar

Soal 13Set 3

Latihan

Nomor 13

Laporkan