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Seja \(f\) uma função tal que: \[ f(3)=-2\quad f'(3)=7\quad f(4)=3\quad f'(4)=-6 \]
Temos informação suficiente para calcular \[ \lim _{x\to 3}f(x)? \] Selecione todas as afirmações corretas
abaixo.
Sim, \(\lim _{x\to 3}f(x) = -2\) porque \(f(3)\) é dado. Não, porque não nos é dado \(f(x)\) para valores de \(x\) próximos de
3. Não, porque \(f\) não é necessariamente contínua em 3. Sim, porque \(f\) é
derivável em 3, e portanto \(f\) é contínua em 3, e portanto \(\lim _{x\to 3}f(x) = f'(3) = 7\). Sim, porque \(f\) é
derivável em 3, e portanto \(f\) é contínua em 3, e portanto \(\lim _{x\to 3}f(x) = f(3) = -2\). Não, porque,
embora \(f\) seja derivável em 3, isso não significa que \(f\) seja contínua em 3.
Determine os limites a seguir ou escreva DNE se eles não existirem.
\begin{align*} \lim _{x\to 3}\frac {f(x) + 2}{x - 3} &= \class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}}\\ \lim _{x\to 3}\frac {f(x) - 2}{x - 3} &= \class{xm-answer}{\cssId{xm-answer-2}{\phantom{\rule{7em}{1.5em}}}}\\ \lim _{x\to 3}\frac {f(x) - f(3)}{x-3} &= \class{xm-answer}{\cssId{xm-answer-3}{\phantom{\rule{7em}{1.5em}}}}\\ \lim _{h\to 1}\frac {f(3+h) + 2}{h} &= \class{xm-answer}{\cssId{xm-answer-4}{\phantom{\rule{7em}{1.5em}}}}\\ \lim _{h\to 0}\frac {f(3+h) + 2}{h} &= \class{xm-answer}{\cssId{xm-answer-5}{\phantom{\rule{7em}{1.5em}}}}\\ \lim _{h\to 0}\frac {f(4+h) - 3}{h} &= \class{xm-answer}{\cssId{xm-answer-6}{\phantom{\rule{7em}{1.5em}}}}\\ \lim _{h\to 0}\frac {f(4+h) - f(4)}{h} &= \class{xm-answer}{\cssId{xm-answer-7}{\phantom{\rule{7em}{1.5em}}}} \end{align*}
A equação da reta tangente à curva \(y = f(x)\) no ponto em que \(x = 3\) é dada por \[ y = \class{xm-answer}{\cssId{xm-answer-8}{\phantom{\rule{7em}{1.5em}}}}. \]