# 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:

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.

• Use the fact that $T(t) /(t-\pi/4)^{2}\to (\cos \pi/4)/2$. Thus the limit is $$\dfrac{2}{\cos(\pi/4)}\lim_{t\to \pi/4}\dfrac{S(t)-T(t)}{(t-\pi/4)^{3}}$$ – Paramanand Singh Feb 25 '17 at 12:07

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 $\dfrac{\tan a} {6}=\dfrac{1}{6}$.