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Question 13

Number 13

The value of $\lim_{x \to -y} \frac{\tan x + \tan y}{\frac{x^2 - y^2}{-2y^2} \cdot (1 - \tan x \tan y)}$ is....

$-1$

$1$

$0$

$y$

$-y$

Explanation

Remember the concept that

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

Then

\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}

Because $x \to -y$ then $x + y \to 0$ , so $\lim_{x + y \to 0} \frac{\tan(x + y)}{x + y} = 1$ . Then

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

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

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

= y

Comments

Question 13Set 3

Exercises

Number 13

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