• Matéria: Matemática
  • Autor: seilamenor
  • Perguntado 5 anos atrás

URGENTEE!!
ENCONTRE O LIMITE:

Anexos:

Respostas

respondido por: adrielxx1
1

Resposta:

Na letra d) pode ter algo errado com a questão. Verifique!

Explicação passo-a-passo:

a)

\lim_{x \to2}\frac{3x^2-6}{2x+4} \\\\ \lim_{x \to2}\frac{3.2^2-6}{2.2+4} \\\\ \lim_{x \to2}\frac{6}{8}=\frac{6}{8}  \\\\ \lim_{x \to2}\frac{3x^2-6}{2x+4} =\frac{6}{8}\\\\

b)

\lim_{x \to4}\frac{sen(4x)}{5x} \\\\4x=u\\x= \frac{u}{4} \\u \to 0\\\\ \lim_{u \to0}\frac{sen(u)}{\frac{5u}{4} } \\\\ \lim_{u \to0}\frac{4}{5} \frac{sen(u)}{u} \\\\ \lim_{u \to0}\frac{4}{5}=\frac{4}{5} \\\\\\ \lim_{x \to0}\frac{sen(4x)}{5x} =\frac{4}{5}\\\\

c)

\lim_{x \to5}\frac{\frac{1}{y}-\frac{1}{5}}{y-5}

\lim_{x \to5}\frac{\frac{5-y}{5y}}{y-5}\\\\\lim_{x \to5}\frac{\frac{-(y-5)}{5y}}{y-5}\\\\\lim_{x \to5}\frac{-1}{5y}\\\\\lim_{x \to5}\frac{-1}{5.5}\\\\\lim_{x \to5}\frac{-1}{25}=\frac{-1}{25}

d)

apenas se for +∞ então essa é a resolução:

\lim_{n \to \infty} (1-3x)^{2x}\\\\ \lim_{n \to \infty} (1-3x)^{2x}=-\infty\\\\

e)

\lim_{x \to5}\frac{x+2}{x-5} \\\\\lim_{x \to5}\frac{5+2}{5-5} \\\\\lim_{x \to5}\frac{7}{0}=\infty \\\\\lim_{x \to5}\frac{x+2}{x-5}=\infty\\

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