L'Hopital's rule does not help with the given limit. Find the limit some other way.
-$\lim _ { x \rightarrow \theta ^ { + } } \frac { \sec x } { \tan x }$
A) 1
B) 0
C) $\infty$
D) $- 1$

Question

Quizplus · Accepted Answer

The given limit can be simplified by recognizing that $\frac{\sec x}{\tan x} = \frac{1}{\sin x}$. As $x$ approaches $\theta^+$ (assuming $\theta = 0$ for the context of trigonometric functions), $\sin x$ approaches 0 from the positive side, making $\frac{1}{\sin x}$ approach $\infty$.

L'Hopital's Rule Does Not Help with the Given Limit lim⁡x→θ+sec⁡xtan⁡x\lim _ { x \rightarrow \theta ^ { + } } \frac { \sec x } { \tan x }limx→θ+​tanxsecx​

L'Hopital's Rule Does Not Help with the Given Limit $\lim _ { x \rightarrow \theta ^ { + } } \frac { \sec x } { \tan x }$