Question

lim(x-senx)/tg³x
x-->0

Answer 1

$L=\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{x-\mathrm{sen\,}x}{\mathrm{tg^{3}\,}x}$


Indeterminação $0/0.$ Vamos aplicar a regra de L'Hospital:

(se o limite abaixo existir, então)

$L=\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\frac{d}{dx}(x-\mathrm{sen\,}x)}{\frac{d}{dx}(\mathrm{tg^{3}\,}x)}\\ \\ \\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{1-\cos x}{3\,\mathrm{tg^{2}\,}x\cdot \frac{d}{dx}(\mathrm{tg\,}x)}\\ \\ \\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{1-\cos x}{3\,\mathrm{tg^{2}\,}x\cdot \sec^{2}x}\\ \\ \\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{(1-\cos x)\cdot (1+\cos x)}{3\,\mathrm{tg^{2}\,}x\cdot \sec^{2}x\cdot (1+\cos x)}\\ \\ \\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{1-\cos^{2}x}{3\,\mathrm{tg^{2}\,}x\cdot \sec^{2}x\cdot (1+\cos x)}\\ \\ \\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\mathrm{sen^{2}\,}x}{3\,\mathrm{tg^{2}\,}x\cdot \sec^{2}x\cdot (1+\cos x)}$

$=\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\mathrm{sen^{2}\,}x}{3\,\frac{\mathrm{sen^{2}}\,x}{\cos^{2}x}\cdot \sec^{2}x\cdot (1+\cos x)}\\ \\ \\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\mathrm{sen^{2}\,}x}{3\,\mathrm{sen^{2}\,}x\cdot \frac{1}{\cos^{2}x}\cdot \sec^{2}x\cdot (1+\cos x)}\\ \\ \\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\mathrm{sen^{2}\,}x}{3\,\mathrm{sen^{2}\,}x\cdot \sec^{2}x\cdot \sec^{2}x\cdot (1+\cos x)}\\ \\ \\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{1}{3\cdot \sec^{4}x\cdot (1+\cos x)}$


Agora que não há mais nenhuma indeterminação, finalmente temos que

$L=\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{1}{3\cdot \sec^{4}x\cdot (1+\cos x)}\\ \\ \\ =\dfrac{1}{3\cdot \sec^{4}0\cdot (1+\cos 0)}\\ \\ \\ =\dfrac{1}{3\cdot 1^{4}\cdot (1+1)}\\ \\ \\ =\dfrac{1}{3\cdot 1\cdot 2}\\ \\ \\ =\dfrac{1}{6}$


$\Rightarrow~\boxed{ \begin{array}{c} \underset{x\to 0}{\mathrm{\ell im}}~\dfrac{x-\mathrm{sen\,}x}{\mathrm{tg^{3}\,}x}=\dfrac{1}{6} \end{array} }$