Question

$\underset{x\to 0}{Lim}\left(\frac{\sec ax-\sec bx}{x^2}\right)$ is equal to

• A. $\frac{a^2-b^2}{2}$
• B. $\frac{b^2-a^2}{2}$
• C. 0
• D. 1

Answer 1

The correct option is A $\frac{a^2-b^2}{2}$

Now,

$\underset{x\to 0}{lim}\left(\frac{\sec ax-\sec bx}{x^2}\right)$ $\frac{0}{0}$ form.

Now using L'Hospital's rule we get,

$=\underset{x\to 0}{lim}\frac{a\sec ax.\tan ax-b\sec bx.\tan bx}{2x}$$\frac{0}{0}$ form.

Now using L'Hospital's rule we get,

$=\underset{x\to 0}{lim}\frac{a^2({\sec }^{3}ax+\sec ax.{\tan }^{2}ax)-b^2({\sec }^{3}bx+\sec bx.{\tan }^{2}bx)}{2}$

$=\frac{a^2-b^2}{2}$.