Question

please answer now please do it fast

Answer 1

Answer:

use the suitable formulae mentioned in the page

Answer 2

On direct application of $\sf{\theta=0,}$ we get indeterminate form.

Then by L'hospital's Rule,

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

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

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