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One-sided infinite limit of tan(x) as x approaches pi/2 from the left

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

Recall that tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \dfrac{\sin(x)}{\cos(x)}, and that cos⁡(x)→0+\cos(x) \to 0^+ as x→(π2)−x \to \left(\dfrac{\pi}{2}\right)^-, while sin⁡(x)→1\sin(x) \to 1. Using this information, evaluate the one-sided limit lim⁡x→(π/2)−tan⁡(x).\lim_{x \to (\pi/2)^-} \tan(x). State the value of the limit (using +∞+\infty or −∞-\infty if appropriate).

Hint

Think about the sign and size of sin⁡(x)\sin(x) and cos⁡(x)\cos(x) as xx approaches π/2\pi/2 from values slightly less than π/2\pi/2. What happens to the fraction sin⁡(x)/cos⁡(x)\sin(x)/\cos(x) when the denominator is a tiny positive number and the numerator is close to 1?

Step-by-step solution

As x→(π/2)−x \to (\pi/2)^-, xx approaches π/2\pi/2 from angles slightly smaller than π/2\pi/2, where cosine is still positive but shrinking toward 0, so cos⁡(x)→0+\cos(x) \to 0^+. Meanwhile sin⁡(x)→sin⁡(π/2)=1\sin(x) \to \sin(\pi/2) = 1, a fixed positive number close to 1.

Since tan⁡(x)=sin⁡(x)/cos⁡(x)\tan(x) = \sin(x)/\cos(x) has a numerator near 11 and a denominator that is a tiny positive number approaching 0, the quotient grows without bound in the positive direction. Therefore lim⁡x→(π/2)−tan⁡(x)=+∞\lim_{x \to (\pi/2)^-} \tan(x) = +\infty, indicating a vertical asymptote at x=π/2x = \pi/2.

Sign of tangent approaching from the right side

Hint
Step-by-step solution

The limit is negative infinity. Just to the right of pi over two, cosine becomes negative while sine stays close to positive one, so sine over cosine becomes a large negative number. As x approaches pi over two from the right, cosine approaches zero from the negative side, making the ratio plunge toward negative infinity. This confirms the vertical asymptote has opposite infinite behavior on each side.

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