Question

please solve it
please please

Answer 1

Answer:

all formulae mentioned , use the limits to suit the formules

Answer 2

(1) Refer https://brainly.in/question/15720183

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(2) Formula applied here is,

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• $\displaystyle\sf {\lim_{x\to0}\dfrac {\sin x}{x}=1}$

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So,

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$\displaystyle\longrightarrow\sf {\lim_{\theta\to0}\dfrac {8\sin\theta-\theta\cos\theta}{3\tan\theta+\theta^2}=\lim_{\theta\to 0}\dfrac {8\sin\theta\cos\theta-\theta\cos^2\theta}{3\sin\theta+\theta^2\cos\theta}}$

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$\displaystyle\longrightarrow\sf {\lim_{\theta\to0}\dfrac {8\sin\theta-\theta\cos\theta}{3\tan\theta+\theta^2}=\dfrac {\displaystyle8\lim_{\theta\to 0}\dfrac {\sin\theta}{\theta}\cdot\theta\cos\theta-\lim_{\theta\to 0}\theta\cos^2\theta}{\displaystyle3\lim_{\theta\to 0}\dfrac {\sin\theta}{\theta}\cdot\theta+\lim_{\theta\to 0}\theta^2\cos\theta}}$

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$\displaystyle\longrightarrow\sf {\lim_{\theta\to0}\dfrac {8\sin\theta-\theta\cos\theta}{3\tan\theta+\theta^2}=\dfrac {\displaystyle\theta\left [8\lim_{\theta\to 0}\dfrac {\sin\theta}{\theta}\cdot\cos\theta-\lim_{\theta\to 0}\cos^2\theta\right]}{\displaystyle\theta\left [3\lim_{\theta\to 0}\dfrac {\sin\theta}{\theta}+\lim_{\theta\to 0}\theta\cos\theta\right]}}$

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$\displaystyle\longrightarrow\sf {\lim_{\theta\to0}\dfrac {8\sin\theta-\theta\cos\theta}{3\tan\theta+\theta^2}=\dfrac {\displaystyle8\lim_{\theta\to 0}\dfrac {\sin\theta}{\theta}\cdot\lim_{\theta\to 0}\cos\theta-\lim_{\theta\to 0}\cos^2\theta}{\displaystyle3\lim_{\theta\to 0}\dfrac {\sin\theta}{\theta}+\lim_{\theta\to 0}\theta\cos\theta}}$

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$\displaystyle\longrightarrow\sf {\lim_{\theta\to0}\dfrac {8\sin\theta-\theta\cos\theta}{3\tan\theta+\theta^2}=\dfrac {8\cdot1\cdot1-1}{3\cdot1}}$

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$\displaystyle\longrightarrow\sf {\underline {\underline {\lim_{\theta\to0}\dfrac {8\sin\theta-\theta\cos\theta}{3\tan\theta+\theta^2}=\dfrac {7}{3}}}}$

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(3) Refer https://brainly.in/question/15721453