Question

Evaluate : ${lim}_{x\to 0}\frac{1+ta{n}^2\sqrt{x}}{2x}$.

Answer 1

$\underset{x\to 0}{lim}\frac{1+{\tan }^2\sqrt{x}}{2x}$

$=\underset{x\to 0}{lim}\frac{{\sec }^2\sqrt{x}}{2x}$

$\underset{x\to 0^{+}}{lim}\frac{{\sec }^2\sqrt{x}}{2x}\to \infty$

$\underset{x\to 0^{-}}{lim}\frac{{\sec }^2\sqrt{x}}{2x}\to -\infty$

$\therefore \underset{x\to 0}{lim}\frac{1+{\tan }^2\sqrt{x}}{2x}$ does not exist.