Evaluate $\lim_{t\to0}\frac{-(t-2)(\sin t +1)-(t+2)\cos t}{(\sin t + \cos t - 1)t^2}$ without L'hopital

Question

Question:

Let \begin{align} &S(t):=\int_{\pi/4}^t (\sin t-\sin\left(\frac\pi4\right))dt\\ &T(t):=\frac{\left(\sin t-\sin\left(\frac\pi4\right)\right)\left(t-\frac\pi4\right)}2\\ \end{align} Using $$\lim_{t\to0}\frac{\tan t - t}{t^3}=\frac13\tag1$$ Evaluate the following (without L'hopital) \begin{align} &\quad\lim_{t\to\frac\pi4}\frac{S(t)-T(t)}{T(t)\left(t-\frac\pi4\right)} \end{align}

What I've done so far is: $$\lim_{t\to\frac\pi4}\frac{S(t)-T(t)}{T(t)\left(t-\frac\pi4\right)}=\lim_{t\to0}\frac{-\cos\left(t+\frac\pi4\right)+\frac{\sqrt2}2-\left(\sin\left(t+\frac\pi4\right)+\frac{\sqrt2}2\right)\frac t2}{\left(\sin\left(t+\frac\pi4\right)-\frac{\sqrt2}2\right)\frac {t^2}2}$$ $$=\lim_{t\to0}\frac{-(t-2)(\sin t +1)-(t+2)\cos t}{(\sin t + \cos t - 1)t^2}$$ But I don't know how to go from here to the Eq.$(1)$. Thanks.

Answer 1

This is an extension of my comment.

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Note that if $a=\pi/4$ (this greatly simplifies typing!!!) then $$S(t) =\cos a - \cos t - (t - a) \sin a, 2T(t)=(t-a)(\sin t - \sin a)$$ therefore $$2S(t)-2T(t)=2(\cos a - \cos t) - (t-a) (\sin a + \sin t) $$ or $$S(t)-T(t)=2\sin\frac{t+a}{2}\sin\frac{t-a}{2}-(t-a)\sin\frac{t+a}{2}\cos\frac{t-a}{2}$$ Let $2u=t+a,2v=t-a$ so that $u\to a, v\to 0$ as $t\to a$. Then we have $$S(t) - T(t) = 2\sin u\cos v(\tan v-v)$$ and hence the desired limit is $$\frac{2}{\cos a} \lim_{t\to a} \frac{S(t) - T(t)} {(t-a) ^{3}}=\frac{1}{2\cos a} \lim_{v\to 0}\sin u\cos v\cdot\frac{\tan v-v}{v^{3}}$$ which is $\dfrac{\tan a} {6}=\dfrac{1}{6}$.

Answer 2

Based on your calculations, the given limit and the sum to product formulae,

\begin{align} & \lim_{t\to\frac\pi4}\frac{S(t)-T(t)}{T(t)\left(t-\frac\pi4\right)} \\ =&\lim_{t\to0}\frac{-\cos\left(t+\frac\pi4\right)+\frac{\sqrt2}2-\left(\sin\left(t+\frac\pi4\right)+\frac{\sqrt2}2\right)\frac t2}{\left(\sin\left(t+\frac\pi4\right)-\frac{\sqrt2}2\right)\frac {t^2}2} \\ =& \lim_{t\to0}\frac{-\cos\left(t+\frac\pi4\right)+\cos(\frac{\pi}{4})-\left(\sin\left(t+\frac\pi4\right)+\sin(\frac{\pi}{4})\right)\frac t2}{\left(\sin\left(t+\frac\pi4\right)-\sin(\frac{\pi}{4})\right)\frac {t^2}2} \\ =& \lim_{t\to0}\frac{-2\sin\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{-t}{2})-\left(2\sin\left(\frac{t}{2}+\frac\pi4\right)\cos(\frac{t}{2})\right)\frac t2}{\left(2\cos\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{t}{2})\right)\frac {t^2}2} \\ =& \lim_{t\to0}\frac{2\sin\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{t}{2})-t\sin\left(\frac{t}{2}+\frac\pi4\right)\cos(\frac{t}{2})} {t^2\cos\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{t}{2})} \\ =& \lim_{t\to0} \frac{\sin(\frac{t}{2} + \frac{\pi}{4})}{\cos(\frac{t}{2} + \frac{\pi}{4})} \cdot \frac{2\sin\left( \frac{t}{2} \right) - t \cos\left( \frac{t}{2} \right)}{t^2 \sin\left( \frac{t}{2} \right)} \\ =& \lim_{t\to0} \tan\left(\frac{t}{2} + \frac{\pi}{4}\right) \cdot \frac{2\sin\left( \frac{t}{2} \right) - t \cos\left( \frac{t}{2} \right)}{t^2 \sin\left( \frac{t}{2} \right)} \\ =& \lim_{t\to0} \tan\left(t + \frac{\pi}{4}\right) \cdot \frac{2\sin\left( t \right) - 2t \cos\left( t \right)}{(2t)^2 \sin\left( t \right)} \\ =& \lim_{t\to0} \tan\left(t + \frac{\pi}{4}\right) \cdot \frac{\tan\left( t \right) - t}{2t^2 \tan\left( t \right)} \\ =& \lim_{t\to0} \frac{1}{2} \tan\left(t + \frac{\pi}{4}\right) \cdot \frac{\tan\left( t \right) - t}{t^3} \frac{t}{\tan\left( t \right)} \\ =& \lim_{t\to0} \frac{1}{2} \tan\left(t + \frac{\pi}{4}\right) \cdot \frac{\tan\left( t \right) - t}{t^3} \frac{t}{\sin\left( t \right)} \cos\left( t \right) \\ =& \frac{1}{2} \tan\left( 0 + \frac{\pi}{4} \right) \cdot \frac{1}{3} \cdot 1 \cdot \cos(0) \\ =& \frac{1}{6} \end{align}