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Evaulate the limit of $\mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)]$.

$$\lim \limits_{\theta \to \frac{\pi}{2}} \mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)] $$

I have attempted the problem by direct substitution and get:

$$\mathbf{tan}^2(\frac{\pi}2)[1-\mathbf{sin}(\frac{\pi}2)] $$ $$\infty^2 \cdot [0]$$ $$\infty \cdot 0$$ I do not know whether to state that it is zero, but I graphed it and got $\frac{1}{2}$. The limit has to be evaluated algebraically.

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    $\begingroup$ seems like others have done the job mostly, I will just notice that $\tan^2 x = \frac{\sin^2 x }{cos^2 x } = \frac{\sin^2 x }{1-\sin^2 x} = \frac{\sin^2 x}{(1-\sin x)(1+\sin x)}$ $\endgroup$ – Alex Aug 21 '16 at 22:08
  • $\begingroup$ @Alex Thanks for another way to do the problem. :) $\endgroup$ – Sigma6RPU Aug 21 '16 at 22:11
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$$\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )[1-{ \sin }(\theta )]=\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )\frac { 1-\sin ^{ 2 }{ (\theta ) } }{ 1+{ \sin }(\theta ) } =\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )\frac { \cos ^{ 2 }{ (\theta ) } }{ 1+{ \sin }(\theta ) } =\\ =\lim _{ \theta \to \frac { \pi }{ 2 } }{ \frac { \sin ^{ 2 }{ (\theta ) } }{ 1+{ \sin }(\theta ) } } =\frac { 1 }{ 2 } $$ Another way is to apply L'hospital's rule $$\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )\left[ 1-\sin { \left( \theta \right) } \right] =\lim _{ \theta \to \frac { \pi }{ 2 } }{ \frac { 1-\sin { \left( \theta \right) } }{ \cot ^{ 2 }{ \left( \theta \right) } } } =\lim _{ \theta \to \frac { \pi }{ 2 } }{ \frac { -\cos { \left( \theta \right) } }{ 2\cot { \left( \theta \right) \left( -\frac { 1 }{ \sin ^{ 2 }{ \left( \theta \right) } } \right) } } } =\frac { 1 }{ 2 } $$

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Hint:

Multiply and divide by $1+\sin\theta$ and simplify.

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  • $\begingroup$ Touché good sir. $\endgroup$ – Sigma6RPU Aug 21 '16 at 21:56
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If you make a substitution: $\sin\theta = \dfrac{x}{1}$ then $\tan\theta =\dfrac{x}{\sqrt{1-x^2}}$.

enter image description here

As $\theta \to \dfrac{\pi}{2}$, we have $\sin\theta = x\to 1$.

$$\lim_{\theta\to\pi/2} \tan^2(\theta)[1-\sin\theta] \ =\ \lim_{x\to 1} \ \dfrac{x^2}{1-x^2}[1-x]$$

$$= \lim_{x\to 1} \ \dfrac{x^2}{(1-x)(1+x)}[1-x] \ = \ \lim_{x\to 1} \ \dfrac{x^2}{(1+x)} = \dfrac{1^2}{1+1} = \dfrac{1}{2}$$

I know this is now what was intended for this exercise, but I thought it would be helpful.

When making substitutions like these you do need to be careful. I chose $\sin\theta$ because it is continuous at $\dfrac{\pi}{2}$.

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$$\lim_{\theta \to \frac{\pi}{2}} (\tan^2\theta) (1-\sin\theta)$$

$$\lim_{\theta \to \frac{\pi}{2}} \left(\dfrac {\sin^2 \theta}{\cos^2{\theta}} \right) (1-\sin\theta)$$

$$\lim_{\theta \to \frac{\pi}{2}} \left(\dfrac {\sin^2 \theta}{1-\sin^2 \theta} \right) (1-\sin\theta)$$

$$\lim_{\theta \to \frac{\pi}{2}} \dfrac {\sin^2 \theta}{(1-\sin \theta)(1+\sin \theta)} (1-\sin\theta)$$

$$\lim_{\theta \to \frac{\pi}{2}} \dfrac {\sin^2 \theta}{1+\sin \theta} $$

Direct substitution now yields $\dfrac 12$.

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Another way to do it.

Just as @zhw. answered $$\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )[1-{ \sin }(\theta )]=\lim_{t\to 0}\, \cot^2(t)(1-\cos (t))$$ Now, use Taylor series built around $t=0$ $$\cos(t)=1-\frac{t^2}{2}+O\left(t^4\right)$$ $$1-\cos(t)=\frac{t^2}{2}+O\left(t^4\right)$$ $$\cot(t)=\frac{1}{t}-\frac{t}{3}-\frac{t^3}{45}+O\left(t^4\right)$$ $$\cot^2(t)=\frac{1}{t^2}-\frac{2}{3}+\frac{t^2}{15}+O\left(t^3\right)$$ $$\cot^2(t)(1-\cos (t))=\left(\frac{t^2}{2}+O\left(t^4\right) \right)\left(\frac{1}{t^2}-\frac{2}{3}+\frac{t^2}{15}+O\left(t^3\right) \right)=\frac{1}{2}-\frac{t^2}{3}+O\left(t^3\right)$$ which shows the limit and how it is approached.

Just for fun, plot on the same graph functions $f(t)=\cot^2(t)(1-\cos (t))$ and $g(t)=\frac{1}{2}-\frac{t^2}{3}$ for $-1 \leq t \leq 1$; you will be surprised to notice how close they are to eachother. Suppose now that you need to solve for $t$ equation $f(t)=\frac 14$. Using $g(t)$ as an approximation will give $t=\pm \frac{\sqrt{3}}{2}\approx \pm 0.866$ while the exact solution (much more difficult to obtain) would be $t=\pm \cos ^{-1}\left(\frac{1}{8} \left(1+\sqrt{17}\right)\right) \approx \pm 0.876$.

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If we let $\theta = \pi/2-t,$ we are looking at

$$\lim_{t\to 0}\, \tan^2(\pi/2-t)(1-\sin(\pi/2-t)) = \lim_{t\to 0}\, \cot^2(t)(1-\cos t).$$

This is the same as

$$\lim_{t\to 0}\,\cos^2t\cdot\frac{t^2}{\sin^2 t}\cdot\frac{1-\cos t}{t^2} = 1\cdot 1^2\cdot (1/2) = 1/2.$$

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