Question

Limite. F(x). 2x-4/5x-15. Z. X=3

Answer 1

$\lim_{x \to \ 3} = \frac{2x-4}{5x-15} \\ \\ \lim_{x \to \ 3} = \frac{2.3-4}{5.3-15} \\ \\ \lim_{x \to \ 3} = \frac{2}{0}$

Vamos calcular os limites laterais

$\lim_{x \to \ ^{-} 3 } = \frac{2x-4}{5x-15} \\ \\ \lim_{x \to \ ^{-} 3 } = \frac{2.29999-4}{5.29999-15} \\ \\ \lim_{x \to \ ^{-} 3 } = \frac{5,9998-4}{14,9995-15} \\ \\ \lim_{x \to \ ^{-} 3 } = \frac{1.9998}{-0,0005} \\ \\ \lim_{x \to \ ^{-} 3 } = -\infty \\ \\ \lim_{x \to \ ^{+} 3 } = \frac{2x-4}{5x-15} \\ \\ \lim_{x \to \ ^{+} 3 } = \frac{2.3.0001-4}{5.30001-15} \\ \\ \lim_{x \to \ ^{+} 3 } = \frac{6,0002-4}{15,0005-15} \\ \\ \lim_{x \to \ ^{+} 3 } = \frac{2,0002}{0,0005}$

$\lim_{x \to \ ^{+} 3 } = +\infty}$

$\lim_{x \to \ 3 } = \frac{2x-4}{5x-15}$

Não existe pois os limites laterais não são iguais.