Evaluate the following limits:

Question:

Evaluate the following limits:

$\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1}$

 

Solution:

$=\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \times \sin x \times \sin x+\sin x-1}{2 \times \sin x \times \sin x-3 \sin x+1}$

$=\lim _{x \rightarrow \frac{\pi}{6}} \frac{(2 \sin x-1) \times(\sin x+1)}{(2 \sin x-1)(\sin x-1)}$

$=-3$

$\therefore \lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \times \sin x \times \sin x+\sin x-1}{2 \times \sin x \times \sin x-3 \sin x+1}=-3$

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now