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  • <img alt="\mathrm{\text { Evaluate } \lim _{t \rightarrow 0} \frac{-(t-2)(\sin t+1)-(t+2) \cos t}{(\sin t+\cos t-1) t^2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Ctext%2

\mathrm{\text { Evaluate } \lim _{t \rightarrow 0} \frac{-(t-2)(\sin t+1)-(t+2) \cos t}{(\sin t+\cos t-1) t^2}}

Option: 1

\frac{1}{6}


Option: 2

0


Option: 3

1


Option: 4

2


Answers (1)

best_answer

Note that if  \mathrm{a=\pi / 4} (this greatly simplifies typing!!!) then

                        \mathrm{ S(t)=\cos a-\cos t-(t-a) \sin a, 2 T(t)=(t-a)(\sin t-\sin a) }
therefore

                     \mathrm{ 2 S(t)-2 T(t)=2(\cos a-\cos t)-(t-a)(\sin a+\sin t) }
or
                       \mathrm{ 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 \mathrm{2 u=t+a, 2 v=t-a} so that \mathrm{u \rightarrow a, v \rightarrow 0} as \mathrm{t \rightarrow a}. Then we have

                           \mathrm{ S(t)-T(t)=2 \sin u \cos v(\tan v-v) }
and hence the desired limit is

                        \mathrm{ \frac{2}{\cos a} \lim _{t \rightarrow a} \frac{S(t)-T(t)}{(t-a)^3}=\frac{1}{2 \cos a} \lim _{v \rightarrow 0} \sin u \cos v \cdot \frac{\tan v-v}{v^3} }

which is \mathrm{\frac{\tan a}{6}=\frac{1}{6}.}

Posted by

HARSH KANKARIA

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