Question

limit x tends to zero sec4x-sec2x/sec3x-secx

Answer 1

Hi dear ❤️,

first change sec terms to cos . We know sec theta= 1/cos theta then solve. There are many methods to solve it but I explain you in simple .so do not confuse .

Step-by-step explanation:see the attachment below .ask doubts in comments. Follow me and mark answer as brainlist.

Answer 2

Answer:

$\lim _{x\to \:0}\left(\frac{\sec \left(4x\right)-\sec \left(2x\right)}{\sec \left(3x\right)-\sec \left(x\right)}\right)=\frac{3}{2}$

Step-by-step explanation:

The given limit is

$\lim _{x\to \:0}\left(\frac{\sec \left(4x\right)-\sec \left(2x\right)}{\sec \left(3x\right)-\sec \left(x\right)}\right)$

Apply the L'Hospital's rule

$\lim _{x\to \:0}\left(\frac{\sec \left(4x\right)\tan \left(4x\right)\cdot \:4-\sec \left(2x\right)\tan \left(2x\right)\cdot \:2}{\sec \left(3x\right)\tan \left(3x\right)\cdot \:3-\sec \left(x\right)\tan \left(x\right)}\right)$

Again apply L'Hospital's rule

$\lim _{x\to \:0}\left(\frac{4\left(4\tan ^2\left(4x\right)\sec \left(4x\right)+4\sec ^3\left(4x\right)\right)-2\left(-2\sec \left(2x\right)+4\sec ^3\left(2x\right)\right)}{9\tan ^2\left(3x\right)\sec \left(3x\right)+9\sec ^3\left(3x\right)+\sec \left(x\right)-2\sec ^3\left(x\right)}\right)$

Plug in the value x = 0

$=\frac{4\left(4\tan ^2\left(4\cdot \:0\right)\sec \left(4\cdot \:0\right)+4\sec ^3\left(4\cdot \:0\right)\right)-2\left(-2\sec \left(2\cdot \:0\right)+4\sec ^3\left(2\cdot \:0\right)\right)}{9\tan ^2\left(3\cdot \:0\right)\sec \left(3\cdot \:0\right)+9\sec ^3\left(3\cdot \:0\right)+\sec \left(0\right)-2\sec ^3\left(0\right)}$

On simplifying, we get

$\lim _{x\to \:0}\left(\frac{\sec \left(4x\right)-\sec \left(2x\right)}{\sec \left(3x\right)-\sec \left(x\right)}\right)=\frac{3}{2}$