Question

Limit x approaches 0 (tan^2x-sin) divide by x4

Answer 1

`\displaystyle \lim_(\theta \to 0)\cfrac{\tan^2(\theta )-\sin(\theta )}{\theta 4}\implies \stackrel{\textit{\Large L'Hopital's rule}}{\lim_(\theta \to 0)\cfrac{(d)/(d\theta )[\tan^2(\theta )-\sin(\theta )]}{(d)/(d\theta )[\theta 4]}} \\\\[-0.35em] ~\dotfill`

`\cfrac{d}{d\theta }[\tan^2(\theta )-\sin(\theta )]\implies \cfrac{d}{d\theta }[[\tan(\theta )]^2-\sin(\theta )]\implies \stackrel{chain~rule}{2\tan(\theta )\cdot \sec^2(\theta )} ~~ -\cos(\theta ) \\\\\\ 2\tan(\theta )\cdot \cfrac{1}{\cos^2(\theta )}-\cos(\theta )\implies \cfrac{~~ (2\sin(\theta) )/(\cos(\theta) ) ~~}{\cos^2(\theta )}~~ -\cos(\theta ) \\\\\\ \cfrac{2\sin(\theta) }{\cos^3(\theta) }~~ -\cos(\theta )\implies \cfrac{2\sin(\theta)-\cos^4(\theta)}{\cos^3(\theta)} \\\\[-0.35em] ~\dotfill`

`\cfrac{d}{d\theta}[4\theta]\implies 4 \\\\[-0.35em] ~\dotfill`

`\displaystyle \lim_(\theta \to 0)\cfrac{\tan^2(\theta )-\sin(\theta )}{\theta 4}\implies \lim_(\theta \to 0)\cfrac{ ~~ (2\sin(\theta)-\cos^4(\theta))/(\cos^3(\theta)) ~~ }{4}\implies \lim_(\theta \to 0)\cfrac{2\sin(\theta)-\cos^4(\theta)}{4\cos^3(\theta)} \\\\\\ \displaystyle \lim_(\theta \to 0)\cfrac{2\sin(0)-\cos^4(0)}{4\cos^3(0)}\implies \lim_(\theta \to 0)\cfrac{2(0)-1^4}{4(1^3)}\implies {\Large \begin{array}{llll} -\cfrac{1}{4} \end{array}}`