Exerc 1 f(x)=\frac{sen(3x)}{9x} x\neq 0
L x=0 f(x)é contínua qual o valor de L?[/tex]
Exerc 2 O lado de um triangulo equilátero está crescendo a uma taxa de 2cm/s.No instante que a área deste triangulo for de
![25\sqrt[]{3}{cm}^{2}](/latexrender/pictures/38a0adfaefec3a69a77f26fce490e625.png "25\sqrt[]{3}{cm}^{2}")
,a taxa de variação da área será(área do triangulo equilatero de lado é s=![\frac{\sqrt[]{3}}{4}/2](/latexrender/pictures/7e0c9c235983351fc761822aded159d1.png "\frac{\sqrt[]{3}}{4}/2")
![{A}_{t.e}=(\sqrt[]{3}.{l}^{2})/2\Rightarrow
A '=(\sqrt[]{3}.2.l.l ' )/2=\sqrt[]{3} . l.l '...](/latexrender/pictures/b417576cb422aed2036cbfea382f0372.png "{A}_{t.e}=(\sqrt[]{3}.{l}^{2})/2\Rightarrow
A '=(\sqrt[]{3}.2.l.l ' )/2=\sqrt[]{3} . l.l '...")
,como ![l ' =2(cm/s)\Rightarrow A '=2.\sqrt[]{3}l](/latexrender/pictures/9cc079b2a15146c343bc6046d982aa92.png "l ' =2(cm/s)\Rightarrow A '=2.\sqrt[]{3}l")
![A=25.\sqrt[]{3}](/latexrender/pictures/bb1f4aa15a1d03c211dab7d73f0f70ef.png "A=25.\sqrt[]{3}")
,teremos:![25.\sqrt[]{3}=(\sqrt[]{3}/2).{l}^{2}\Rightarrow l=5.\sqrt[]{2}cm](/latexrender/pictures/411df85a6b7bcf8af595c25d4e84d90f.png "25.\sqrt[]{3}=(\sqrt[]{3}/2).{l}^{2}\Rightarrow l=5.\sqrt[]{2}cm")
, entao:![A ' =2.\sqrt[]{3}.l=2.\sqrt[]{3}.5\sqrt[]{2}=10.\sqrt[]{6}cm2](/latexrender/pictures/627fc8db610a224fb466073f640887c2.png "A ' =2.\sqrt[]{3}.l=2.\sqrt[]{3}.5\sqrt[]{2}=10.\sqrt[]{6}cm2")
por ![\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}](/latexrender/pictures/981987c7bcdf9f8f498ca4605785636a.png "\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}")
![\sqrt[]{2}](/latexrender/pictures/f21662d1cabab6e8b273a4b6f1cd663a.png "\sqrt[]{2}")
e elevar ao quadrado os dois lados)