[tex]\displaystyle \sf \text{Fa\c camos} : \\\\ \lim_{x\to 0} \frac{\sqrt{3+sen(x)}-\sqrt{3-tg(x)}}{x^5} = A \\\\\\ \lim_{x\to 0} \frac{\sqrt{x}-\sqrt[3]{\sf x}}{x^2+3} = B \\\\\\ \text{Resolvendo os limites} : \\\\ A) \\\\ \lim_{x\to 0} \frac{\sqrt{3+sen(x)}-\sqrt{3-tg(x)}}{x^5} \\\\\\ \text{x = 0 dar\'a indtermina\c c\~ao, ent\~ao vamos derivar o numerador e denominador} :[/tex]
[tex]\displaystyle \sf \lim_{x\to 0} \frac{\displaystyle \frac{cos(x)}{2\sqrt{3+sen(x)}}+\frac{sec^2(x)}{2\sqrt{3-tg(x)}}}{5.x^4} = \frac{\displaystyle \sf \frac{cos(0)}{2\sqrt{3+\sen(0)}} + \frac{sec^2(0)}{2\sqrt{3-tg(0)}}}{5.0^4} \\\\\\ A = \frac{\displaystyle \sf \frac{1}{\sqrt{3}}}{0} \to \boxed{\sf A = \infty }[/tex]
[tex]\displaystyle \sf B) \\\\ \lim_{x\to 0} \frac{\sqrt{x}-\sqrt[3]{\sf x}}{x^2+3} \to \lim_{x\to 0 }\frac{\displaystyle \frac{x^{\frac{1}{2}}}{x^2}+\frac{x^{\frac{3}{2}}}{x^2}}{\displaystyle 1 +\frac{3}{x^2}} \\\\\\\[/tex]
[tex]\displaystyle \sf \lim_{x\to 0}\frac{\displaystyle \left(\frac{1}{x}\right)^{\displaystyle \frac{3}{2}}+\left(\frac{1}{x}\right)^{\displaystyle \frac{5}{3}}}{\displaystyle 1 +\frac{3}{x^2}} \to \frac{\displaystyle \underbrace{\left(\frac{1}{\infty}\right)^{\displaystyle \frac{3}{2}}}_{0}+\underbrace{\left(\frac{1}{\infty}\right)^{\displaystyle \frac{5}{3}} }_{0} }{\displaystyle 1 +\underbrace{\frac{3}{\infty ^2}}_{0} } = 0 \\\\\ \boxed{\sf B = 0 }[/tex]
[tex]\displaystyle \sf \text{Temos : } \\\\ \int^{A}_B \frac{16}{8+4x^2}dx \to \int^{\infty}_0\frac{16}{4(2+x^2)}dx \to 4\int^{\infty}_0\frac{1}{2+x^2}dx\\\\\\ \lim_{M\to \infty}4\int ^{M}_0 \frac{1}{\displaystyle \sf 2\left(1+\frac{x^2}{2}\right)}\to \lim_{M\to \infty} \frac{2}{\sqrt{2}}\int ^{M}_0 \frac{\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle 1+\left(\frac{x}{\sqrt{2}}\right)^2}dx[/tex]
[tex]\displaystyle \sf \text{Fa\c camos} : \\\\\ u = \frac{x}{\sqrt{2}} \to du =\frac{1}{\sqrt{2}}dx \\\\ Da{\'i}}: \\\\ \lim_{M\to \infty} \frac{2}{\sqrt{2}} \int^{M}_{0} \frac{1}{1+u^2}du \to \frac{2}{\sqrt{2}} \lim_{M\to \infty} \left. Arc \ tg\left(\frac{x}{\sqrt{2}}\right)\right|^M_0 \\\\\\ \frac{2}{\sqrt{2}}\lim_{M\to \infty} \left[Arc\ tg\left(\frac{M}{\sqrt{2}}\right)-\underbrace{\sf Arc\ tg\left(\frac{0}{\sqrt{2}}\right)}_{0} \right][/tex]
[tex]\displaystyle \sf \frac{2}{\sqrt{2}}\ \ \underbrace{\sf \lim_{M\to \infty}\ Arc\ tg\left(\frac{M}{\sqrt{2}}\right) }_{\displaystyle \frac{\pi}{2}} \to \frac{2}{\sqrt{2}}\cdot \frac{\pi}{2} \\\\\\ \Large\boxed{\sf \ \frac{\pi }{\sqrt{2}}\ }\checkmark[/tex]
se eu nao errei nenhuma conta é isso aí
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[tex]\displaystyle \sf \text{Fa\c camos} : \\\\ \lim_{x\to 0} \frac{\sqrt{3+sen(x)}-\sqrt{3-tg(x)}}{x^5} = A \\\\\\ \lim_{x\to 0} \frac{\sqrt{x}-\sqrt[3]{\sf x}}{x^2+3} = B \\\\\\ \text{Resolvendo os limites} : \\\\ A) \\\\ \lim_{x\to 0} \frac{\sqrt{3+sen(x)}-\sqrt{3-tg(x)}}{x^5} \\\\\\ \text{x = 0 dar\'a indtermina\c c\~ao, ent\~ao vamos derivar o numerador e denominador} :[/tex]
[tex]\displaystyle \sf \lim_{x\to 0} \frac{\displaystyle \frac{cos(x)}{2\sqrt{3+sen(x)}}+\frac{sec^2(x)}{2\sqrt{3-tg(x)}}}{5.x^4} = \frac{\displaystyle \sf \frac{cos(0)}{2\sqrt{3+\sen(0)}} + \frac{sec^2(0)}{2\sqrt{3-tg(0)}}}{5.0^4} \\\\\\ A = \frac{\displaystyle \sf \frac{1}{\sqrt{3}}}{0} \to \boxed{\sf A = \infty }[/tex]
[tex]\displaystyle \sf B) \\\\ \lim_{x\to 0} \frac{\sqrt{x}-\sqrt[3]{\sf x}}{x^2+3} \to \lim_{x\to 0 }\frac{\displaystyle \frac{x^{\frac{1}{2}}}{x^2}+\frac{x^{\frac{3}{2}}}{x^2}}{\displaystyle 1 +\frac{3}{x^2}} \\\\\\\[/tex]
[tex]\displaystyle \sf \lim_{x\to 0}\frac{\displaystyle \left(\frac{1}{x}\right)^{\displaystyle \frac{3}{2}}+\left(\frac{1}{x}\right)^{\displaystyle \frac{5}{3}}}{\displaystyle 1 +\frac{3}{x^2}} \to \frac{\displaystyle \underbrace{\left(\frac{1}{\infty}\right)^{\displaystyle \frac{3}{2}}}_{0}+\underbrace{\left(\frac{1}{\infty}\right)^{\displaystyle \frac{5}{3}} }_{0} }{\displaystyle 1 +\underbrace{\frac{3}{\infty ^2}}_{0} } = 0 \\\\\ \boxed{\sf B = 0 }[/tex]
[tex]\displaystyle \sf \text{Temos : } \\\\ \int^{A}_B \frac{16}{8+4x^2}dx \to \int^{\infty}_0\frac{16}{4(2+x^2)}dx \to 4\int^{\infty}_0\frac{1}{2+x^2}dx\\\\\\ \lim_{M\to \infty}4\int ^{M}_0 \frac{1}{\displaystyle \sf 2\left(1+\frac{x^2}{2}\right)}\to \lim_{M\to \infty} \frac{2}{\sqrt{2}}\int ^{M}_0 \frac{\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle 1+\left(\frac{x}{\sqrt{2}}\right)^2}dx[/tex]
[tex]\displaystyle \sf \text{Fa\c camos} : \\\\\ u = \frac{x}{\sqrt{2}} \to du =\frac{1}{\sqrt{2}}dx \\\\ Da{\'i}}: \\\\ \lim_{M\to \infty} \frac{2}{\sqrt{2}} \int^{M}_{0} \frac{1}{1+u^2}du \to \frac{2}{\sqrt{2}} \lim_{M\to \infty} \left. Arc \ tg\left(\frac{x}{\sqrt{2}}\right)\right|^M_0 \\\\\\ \frac{2}{\sqrt{2}}\lim_{M\to \infty} \left[Arc\ tg\left(\frac{M}{\sqrt{2}}\right)-\underbrace{\sf Arc\ tg\left(\frac{0}{\sqrt{2}}\right)}_{0} \right][/tex]
[tex]\displaystyle \sf \frac{2}{\sqrt{2}}\ \ \underbrace{\sf \lim_{M\to \infty}\ Arc\ tg\left(\frac{M}{\sqrt{2}}\right) }_{\displaystyle \frac{\pi}{2}} \to \frac{2}{\sqrt{2}}\cdot \frac{\pi}{2} \\\\\\ \Large\boxed{\sf \ \frac{\pi }{\sqrt{2}}\ }\checkmark[/tex]
se eu nao errei nenhuma conta é isso aí