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Question 11

Number 11

The value of $\lim_{x \to \frac{\pi}{2}} \frac{\sec 2x + 2}{\tan 2x}$ is....

$-2$

$-1$

$-\frac{1}{2}$

$0$

$1$

Explanation

Let $y = x - \frac{\pi}{2}$ then $x = y + \frac{\pi}{2}$ . Solve the limit with the above substitution.

= \lim_{y \to 0} \frac{\sec 2\left(y + \frac{\pi}{2}\right) + 2}{\tan 2\left(y + \frac{\pi}{2}\right)}

= \lim_{y \to 0} \frac{-\sec 2y + 2}{\tan 2y}

= \lim_{y \to 0} \frac{-\frac{1}{\cos 2y} + 2}{\frac{\sin 2y}{\cos 2y}}

= \lim_{y \to 0} \frac{\frac{-1 + 2 \cos 2y}{\cos 2y}}{\frac{\sin 2y}{\cos 2y}}

= \lim_{y \to 0} \frac{-1 + 2 \cos 2y}{\sin 2y}

Use L'Hôpital's rule

= \lim_{y \to 0} \frac{-4 \sin 2y}{2 \cos 2y}

= \lim_{y \to 0} -2 \frac{\sin 2y}{\cos 2y}

= \lim_{y \to 0} -2 \tan 2y

= -2 \tan(2 \cdot 0)

= -2 \tan 0

= -2 \cdot 0

= 0

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Question 11Set 3

Exercises

Number 11

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