Advertisement Remove all ads

Lim X → 0 Sec 5 X − Sec 3 X Sec 3 X − Sec X - Mathematics

\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]

Advertisement Remove all ads

Solution

\[\lim_{x \to 0} \left[ \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x} \right]\]

\[= \lim_{x \to 0} \left[ \frac{\frac{1}{\cos 5x}\frac{1}{\cos 3x}}{\frac{1}{\cos 3x} - \frac{1}{\cos x}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\cos 3x - \cos 5x}{\cos 5x \cos 3x\left\{ \frac{\cos x - \cos 3x}{\cos x \cos 3x} \right\}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( \cos3x - \cos5x \right)\cos x}{\cos 5x\left\{ \cos x - \cos3x \right\}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{- 2\sin\left( \frac{3x + 5x}{2} \right)\sin\left( \frac{3x - 5x}{2} \right) \times \cos x}{\cos\left( 5x \right)\left[ - 2\sin\left( \frac{x + 3x}{2} \right)\sin\left( \frac{x - 3x}{2} \right) \right]} \right] \left[ \because cosC - cosD = - 2\sin\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right) \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin\left( 4x \right) \times \sin\left( - x \right) \times \cos x}{\cos\left( 5x \right) \times \sin\left( 2x \right) \times \sin\left( - x \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin\left( 4x \right)}{4x} \times \frac{4x}{\frac{\sin\left( 2x \right)}{2x} \times 2x} \times \frac{\cos x}{\cos5x} \right]\]
\[ = \frac{4}{2}\frac{\cos0}{\cos0}\]
\[ = 2\]
\[\]

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.7 | Q 29 | Page 50
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×