Q.[tex] \\ [/tex][tex] \tt \lim_{ x\to y} \: \frac{tan \: x \: - \: tan \: y}{{((1 \:

$\begin{aligned}&{\lim_{x\to y}}\:\frac{\tan x-\tan y}{\left(1-\frac{x}{y}\right)(\tan x\tan y+1)}\\&{=\ }\bf{-}y\end{aligned}$

Pembahasan

Limit

$\begin{aligned}&{\lim_{x\to y}}\:\frac{\tan x-\tan y}{\left(1-\frac{x}{y}\right)(\tan x\tan y+1)}\\&{=\ }{\lim_{x\to y}}\:\frac{\tan x-\tan y}{\left(\frac{y-x}{y}\right)(\tan x\tan y+1)}\\&{=\ }{\lim_{x\to y}}\:\left(\frac{y}{y-x}\cdot\frac{\tan x-\tan y}{1+\tan x\tan y}\right)\\&{\ \rightsquigarrow}\left[\ \frac{\tan x-\tan y}{1+\tan x\tan y}=\tan(x-y)\right.\\&{=\ }{\lim_{x\to y}}\:\frac{y\tan(x-y)}{y-x}\\&{=\ }{\lim_{x\to y}}\:\left(y\cdot\frac{\tan(x-y)}{y-x}\right)\end{aligned}$
$\begin{aligned}&{\ \rightsquigarrow}\left[\ \begin{aligned}&\textsf{Karena $x\to y$, maka}\\&y\textsf{ adalah konstanta.}\\&\textsf{Keluarkan dari dalam limit}\\&\textsf{menjadi pengali.}\end{aligned}\right.\\&{=\ }y\cdot{\lim_{x\to y}}\:\frac{\tan(x-y)}{y-x}\\&{\ \rightsquigarrow}\left[\ \begin{aligned}&\textsf{Gunakan aturan L'H\^opital}\\&\textsf{karena memenuhi bentuk 0/0.}\\\end{aligned}\right.\end{aligned}$
$\begin{aligned}&{=\ }y\cdot{\lim_{x\to y}}\:\frac{\frac{d}{dx}\tan(x-y)}{\frac{d}{dx}(y-x)}\\&{=\ }y\cdot{\lim_{x\to y}}\:\frac{\sec^2(x-y)\frac{d}{dx}(x-y)}{\frac{d}{dx}(y-x)}\\&{=\ }y\cdot{\lim_{x\to y}}\:\left(\frac{\frac{d}{dx}(x-y)}{\frac{d}{dx}(y-x)}\cdot\sec^2(x-y)\right)\\&{=\ }y\cdot{\lim_{x\to y}}\:\left(\frac{\frac{d}{dx}\left(-(y-x)\right)}{\frac{d}{dx}(y-x)}\cdot\sec^2(x-y)\right)\end{aligned}$
$\begin{aligned}&{=\ }y\cdot{\lim_{x\to y}}\:\left(\frac{-\cancel{\frac{d}{dx}(y-x)}}{\cancel{\frac{d}{dx}(y-x)}}\cdot\sec^2(x-y)\right)\\&{=\ }y\cdot{\lim_{x\to y}}\:\left(-\sec^2(x-y)\right)\\&{=\ }-y\cdot{\lim_{x\to y}}\:\sec^2(x-y)\\&{=\ }-y\cdot{\lim_{x\to y}}\:\frac{1}{\cos^2(x-y)}\end{aligned}$
$\begin{aligned}&{\ \rightsquigarrow}\textsf{ substitusi $x$ dengan $y$}\\&{=\ }-y\cdot\frac{1}{\cos^2(y-y)}\\&{=\ }-y\cdot\frac{1}{\cos^2(0)}\\&{=\ }-y\cdot\frac{1}{1}\\&{=\ }\boxed{\bf{-}y}\\&\blacksquare\end{aligned}$