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One-sided infinite limit of sec(x)

Consider the function f(x)=sec⁡(x)f(x) = \sec(x) near x=π2x = \dfrac{\pi}{2}.

Recall that sec⁡(x)=1cos⁡(x)\sec(x) = \dfrac{1}{\cos(x)}. Evaluate the one-sided limit lim⁡x→π2+sec⁡(x).\lim_{x \to \frac{\pi}{2}^+} \sec(x). In your answer, state whether the limit is +∞+\infty, −∞-\infty, or that it does not exist in any of those senses, and briefly justify your conclusion using the sign and behavior of cos⁡(x)\cos(x) as xx approaches π2\frac{\pi}{2} from the right.

Hint

Rewrite sec⁡(x)\sec(x) as 1cos⁡(x)\dfrac{1}{\cos(x)}, and think about the sign of cos⁡(x)\cos(x) for xx slightly greater than π/2\pi/2, along with how close cos⁡(x)\cos(x) gets to 0.

Step-by-step solution

Since sec⁡(x)=1cos⁡(x)\sec(x) = \dfrac{1}{\cos(x)}, we need to understand the behavior of cos⁡(x)\cos(x) as x→π2+x \to \frac{\pi}{2}^+. Note that cos⁡(π2)=0\cos(\frac{\pi}{2}) = 0. For xx slightly greater than π/2\pi/2, we have cos⁡(x)<0\cos(x) < 0 (since the cosine function is decreasing at x=π/2x = \pi/2). Therefore, as x→π2+x \to \frac{\pi}{2}^+, we have cos⁡(x)→0−\cos(x) \to 0^-, meaning cos⁡(x)\cos(x) approaches 00 from the negative side.

Dividing 11 by a small negative number produces a large negative number, and as the denominator shrinks toward 0−0^-, the quotient grows without bound in the negative direction. Therefore lim⁡x→π2+sec⁡(x)=−∞\lim_{x \to \frac{\pi}{2}^+} \sec(x) = -\infty, indicating a vertical asymptote at x=π2x = \frac{\pi}{2}.

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