A one-sided limit

Calculus Level 3

Find

lim ⁡ t → π 2 − d d t 1 2 − 1 2 sin ⁡ t \large \lim_{t \to \frac{\pi}{2}^- } \dfrac{d}{dt} \sqrt{ \frac{1}{2} - \frac{1}{2} \sin t }

Note: t t is approaching π 2 \frac{\pi}{2} from the left.


The answer is -0.5.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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3 solutions

Chris Lewis
Aug 17, 2020

Let u = π 2 − t u=\frac{\pi}{2}-t . We want the limit as u → 0 + u \to 0^+ .

Then 1 2 − 1 2 sin ⁡ t = 1 2 − 1 2 cos ⁡ u ≈ 1 2 − 1 2 ( 1 − 1 2 u 2 ) (small angle approximation for cosine) = 1 4 u 2 = 1 2 u = π 4 − 1 2 t \begin{aligned} \sqrt{\frac12-\frac12 \sin t} &= \sqrt{\frac12-\frac12 \cos u}\\ & \approx \sqrt{\frac12-\frac12 \left(1-\frac12 u^2 \right)} \;\;\;\text{(small angle approximation for cosine)} \\ &=\sqrt{\frac14 u^2} \\ &=\frac12 u \\ &=\frac{\pi}{4}-\frac12 t \end{aligned}

Differentiating, we want the limit as t → π 2 − t \to \frac{\pi}{2}^- of − 1 2 -\frac12 , which is just − 1 2 \boxed{-\frac12} .

8 Helpful 2 Interesting 6 Brilliant 1 Confused
Chew-Seong Cheong
Aug 18, 2020

L = lim ⁡ t → π 2 − d d t 1 2 − 1 2 sin ⁡ t = lim ⁡ t → π 2 − − cos ⁡ t 4 1 2 − 1 2 sin ⁡ t = lim ⁡ t → π 2 − − 1 − sin ⁡ 2 t 2 2 1 − sin ⁡ t Note that lim ⁡ t → π 2 − cos ⁡ t > 0 = lim ⁡ t → π 2 − − 1 + sin ⁡ t 2 2 = − 1 2 = − 0.5 \begin{aligned} L & = \lim_{t \to \frac \pi 2^-} \frac d{dt} \sqrt{\frac 12 - \frac 12 \sin t} \\ & = \lim_{t \to \frac \pi 2^-} \frac {-\cos t}{4\sqrt{\frac 12 - \frac 12 \sin t}} \\ & = \lim_{t \to \frac \pi 2^-} \frac {-\sqrt{1-\sin^2 t}}{2\sqrt 2 \sqrt{1-\sin t}} & \small \blue{\text{Note that }\lim_{t \to \frac \pi 2^-} \cos t > 0} \\ & = \lim_{t \to \frac \pi 2^-} \frac {-\sqrt{1+\sin t}}{2\sqrt 2} \\ & = - \frac 12 = \boxed{-0.5} \end{aligned}

5 Helpful 0 Interesting 2 Brilliant 0 Confused
Saúl Huerta
Aug 18, 2020

d d t 1 2 − 1 2 sin ⁡ t = − 2 4 ( cos ⁡ t 1 − sin ⁡ t ) \frac{d}{dt} \sqrt{\frac{1}{2}-\frac{1}{2}\sin t}=-\frac{\sqrt{2}}{4}\left(\frac{\cos t}{\sqrt{1-\sin t}}\right) cos ⁡ t 1 − sin ⁡ t = cos ⁡ t 1 − sin ⁡ t ⋅ 1 + sin ⁡ t 1 + sin ⁡ t = cos ⁡ t ⋅ 1 + sin ⁡ t 1 − sin ⁡ 2 t = cos ⁡ t ⋅ 1 + sin ⁡ t cos ⁡ 2 t = 1 + sin ⁡ t \frac{\cos t}{\sqrt{1-\sin t}}=\frac{\cos t}{\sqrt{1-\sin t}}\cdot\frac{\sqrt{1+\sin t}}{\sqrt{1+\sin t}}=\frac{\cos t \cdot \sqrt{1+\sin t}}{\sqrt{1-\sin^2 t}}=\frac{\cos t \cdot \sqrt{1+\sin t}}{\sqrt{\cos^2 t}}=\sqrt{1+\sin t} ⟹ lim ⁡ t → π 2 − d d t 1 2 − 1 2 sin ⁡ t \implies\lim_{t\rightarrow\frac{\pi}{2}^{-}}\frac{d}{dt} \sqrt{\frac{1}{2}-\frac{1}{2}\sin t} = lim ⁡ t → π 2 − − 2 4 1 + sin ⁡ t =\lim_{t\rightarrow\frac{\pi}{2}^{-}} -\frac{\sqrt{2}}{4}\sqrt{1+\sin t} = − 1 2 =\boxed{-\frac{1}{2}}

0 Helpful 0 Interesting 1 Brilliant 0 Confused

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