First slide

Introduction to limits

Question

$lim_{x \to π / 6} \frac{2 \sin^{2} x + \sin x - 1}{2 \sin^{2} x - 3 \sin x + 1} =$

Moderate

A

3
B

-3
C

6
D

0

Solution

We have,
$\begin{array}{l} lim_{x \to π / 6} \frac{2 \sin^{2} x + \sin x - 1}{2 \sin^{2} x - 3 \sin x + 1} \\ = lim_{x \to π / 6} \frac{(2 \sin x - 1) (\sin x + 1)}{(2 \sin x - 1) (\sin x - 1)} \\ = lim_{x \to π / 6} \frac{\sin x + 1}{\sin x - 1} = - 3 \end{array}$

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