Observe o gráfico de f e marque as alternativas corretas.
\(\displaystyle\lim_{x \to -2} f(x)=3\)
\(\displaystyle\lim_{x \to 2^+} f(x)=0\)
\(\displaystyle\lim_{x \to 0^+} f(x)=2\)
\(\displaystyle\lim_{x \to -2^+} f(x)=0\)
\(\displaystyle\lim_{x \to 2} f(x)= \nexists \)
\(\displaystyle\lim_{x \to -2^-} f(x)=3\)
\(\displaystyle\lim_{x \to -2} f(x)= \nexists \)
Seja f uma função tal que f(x) = 3x+2, x \( \in \mathbb{R}\). Se \(\displaystyle\lim_{x \to 1} f(x) = 5\),\(\\\) encontre um \(\delta\) para \(\epsilon\)=0,01, tal que 0 < \(\mid\)x-1\(\mid\) < \(\delta\)\(\rightarrow\)\(\mid\)f(x)-5\(\mid\)< 0,01.
0<\(\delta\)<0,01
0<\(\delta\)< \(\frac{0,01}{5}\)
0<\(\delta\)< \(\frac{0,01}{3}\)
0<\(\delta\)< \(\frac{1}{5}\)
Dada f(x) = \(\frac{\mid x \mid}{x}\). O que podemos afirmar sobre \(\displaystyle\lim_{x \to 0} f(x)\)?
\(\displaystyle\lim_{x \to 0^+} f(x)=0\)
\(\displaystyle\lim_{x \to 0^-} f(x)=-1\)
\(\displaystyle\lim_{x \to 0^+} f(x)=1\)
\(\displaystyle\lim_{x \to 0} f(x)= \nexists\)
Encontre \(\displaystyle\lim_{x \to 1} f(x)\), sabendo que: \[ f(x) = \left\{\begin{array}{rll} 3x-2 & \hbox{se} & x>1 \\ 2 & \hbox{se} & x=0 \\ 4x+1 & \hbox{se} & x<1 \end{array}\right.\]
\(\displaystyle\lim_{x \to 1} f(x)=1\)
\(\displaystyle\lim_{x \to 1} f(x)=5\)
\(\displaystyle\lim_{x \to 1} f(x)=2\)
\(\displaystyle\lim_{x \to 1} f(x)= \nexists\)
Por favor, aguarde! Carregando...