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Question 14

Number 14

If $b, c \neq 0$ and $\lim_{x \to a} \frac{(x - a) \tan(b(a - x))}{\cos(c(x - a)) - 1} = d$ , then $b = ....$

$2c^2d$

$c^2d$

$\frac{1}{2}c^2d$

$-\frac{1}{2}c^2d$

$-c^2d$

Explanation

Change the denominator to

\lim_{x \to a} \frac{(x - a) \tan(b(a - x))}{\cos(c(x - a)) - 1} = d

\lim_{x \to a} \frac{(x - a) \tan(b(a - x))}{-2 \sin^2\left(\frac{1}{2}c(x - a)\right)} = d

\lim_{x \to a} \frac{x - a}{-2 \sin\left(\frac{1}{2}c(x - a)\right)} \cdot \frac{\tan(b(a - x))}{\sin\left(\frac{1}{2}c(x - a)\right)} = d

Because $x \to a$ , then $x - a \to 0$ and $a - x \to 0$ . Using limit properties $\lim_{u \to 0} \frac{\sin u}{u} = 1$ and $\lim_{u \to 0} \frac{\tan u}{u} = 1$

\frac{1}{-2 \cdot \frac{1}{2}c} \cdot \frac{b(-1)}{\frac{1}{2}c} = d

\frac{1}{-c} \cdot \frac{-b}{\frac{1}{2}c} = d

\frac{b}{\frac{1}{2}c^2} = d

\frac{2b}{c^2} = d

2b = c^2d

b = \frac{1}{2}c^2d

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Question 14Set 3

Exercises

Number 14

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