Question

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan x}}{{{x^2} - x}}} \right)\) is equal to ________

Answer 1

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan x}}{{{x^2} - x}}} \right)\)

0/0 form so we can apply L-Hospital’s Rule

\(= \mathop {\lim }\limits_{x \to 0} \frac{{{{\sec }^2}x}}{{2x - 1}} = \frac{1}{{ - 1}} = - 1\)