Matemática, perguntado por diamantefrancis22, 8 meses atrás

Encontre os limites abaixo:


A) lim x-> 0 sen3x / 2x
B) lim x-> 0 senx/ 4x
C) lim x-> 0 tg2x/3x
D) lim x-> 0 sen4x/sen3x
E) lim x-> 0 tg3x/tg5x
F) lim x-> infinito (1+1/x) ^2
G) lim x-> infinito (1+1/x)^ x/3
H) lim x-> infinito (1+1/x)^x+2
I) lim x-> infinito (1+4/x)^x
J) lim x-> infinito (1 - 2/x) ^3x

Soluções para a tarefa

Respondido por RoRoHoul
0

A)

\displaystyle \lim_{x \to 0} \frac{sen3x}{2x}  = lim_{x \to 0} \frac{\frac{3}{2}sen3x}{3x} = \frac{3}{2}1=\frac{3}{2}

B)

\displaystyle lim_{x \to 0} \frac{senx}{4x} =  \lim_{x \to 0} \frac{\frac{1}{4}senx}{x}=\frac{1}{4}1=\frac{1}{4}

C)

\displaystyle lim_{x \to 0} \frac{tg2x}{3x} = lim_{x \to 0} \frac{\frac{sen2x}{cos2x}}{3x}= lim_{x \to 0} \frac{sen2x}{3xcos2x} =  lim_{x \to 0} \frac{\frac{2}{3}sen2x}{2xcos2x} = \frac{2}{3cos0}=\frac{2}{3}

D)

\displaystyle  lim_{x \to 0} \frac{sen4x}{sen3x} = lim_{x \to 0} \frac{\cos(4x)\cdot 4}{\cos(3x)\cdot \:3} = \frac{4}{3}

E)

\displaystyle \lim_{x \to 0} \frac{tg3x}{tg5x} = \frac{3}{5}

F)

\displaystyle \lim_{x \to \infty} \left(1+\frac{1}{x}\right)^2=(1+0)^2=1

G)

\displaystyle \lim_{x \to \infty} \left(1+\frac{1}{x}\right)^\frac{x}{3}=\lim_{x \to \infty} \sqrt[3]{\left(1+\frac{1}{x}\right)^x}=\sqrt[3]{e}

H)

\displaystyle \lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x+2=e+2

I)

\displaystyle \lim_{x \to \infty} \left(1+\frac{4}{x}\right)^x = e^4

J)

\displaystyle \lim_{x \to \infty} \left(1-\frac{2}{x}\right)^{3x}=\frac{1}{e^{2.3}}=\frac{1}{e^6}

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