Cp1 :2005
作者:
Antônio Marcos Barbosa
最近上传:
9 年前
许可:
Creative Commons CC BY 4.0
摘要:
Primeiras questões respondidas do banco de Cálculo 1 da UFAL.
\begin
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Primeiras questões respondidas do banco de Cálculo 1 da UFAL.
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
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\begin{document}
\section{$1^a$ avalia\c{c}\~ao 21/02/2005}
\begin{enumerate}
\item Determine as ass\'intotas verticais e horizontais, caso existam, ao gr\'afico da fun\c{c}\~ao $$f(x)=\dfrac{x^{3}-1}{5x^{3}-20x^{2}+15x}.$$
Resposta: \\
i) Raí\'izes de $5x^{3}-20x^{2}+15$
\begin{align*}
5x^{3}-20x^{2}+15&=0\\
5x(x^2-4x+3)&=0
\end{align*}
$x=0 \mbox{ ou } x^2-4x+3=0$ donde $x'=1, x''=3$
ii) ra\'izes de $x^{3}-1$
$$x^{3}-1=0 \mbox{ donde }x=1$$
logo, $$x^3-1=(x-1)(x^2+x+1)$$
Que tem como ra\'izes apenas $x=1$.
Analisar:
\begin{align}
\lim_{x \rightarrow 0} f(x)&=\lim_{x \rightarrow 0} \dfrac{(x^2+x+1)(x-1)}{5x(x-3)(x-1)}\\
&=\lim_{x \rightarrow 0} \dfrac{(x^2+x+1)}{5x(x-3)}=+\infty
\end{align}
$x=0$ \'e ass\'intota vertical.
\begin{align}
\lim_{x \rightarrow 1} f(x)&=\lim_{x \rightarrow 1} \dfrac{(x^2+x+1)}{5x(x-3)}=-\dfrac{3}{10}
\end{align}
\begin{align}
\lim_{x \rightarrow 3} f(x)&=\lim_{x \rightarrow 3} \dfrac{(x^2+x+1)}{5x(x-3)}=+ \infty
\end{align}
$x=3$ é ass\'intota vertical.
\begin{align}
\lim_{x \rightarrow +\infty} f(x)&=\lim_{x \rightarrow +\infty} \dfrac{(x^2+x+1)}{5x(x-3)}\\
&=\lim_{x \rightarrow +\infty} \dfrac{(x^2+x+1)}{5x^2-15x}\\
&=\lim_{x \rightarrow +\infty} \dfrac{1+\frac{1}{x}+\frac{1}{x^2}}{5-\frac{15}{x}}=\frac{1}{5}
\end{align}
Note que $\lim_{x \rightarrow +\infty} f(x)=\lim_{x \rightarrow -\infty} f(x)$ logo $y=\frac{1}{5}$ \'e ass\'intota vertical.
\item Calcule os seguintes limites:
\vspace{0.2cm}
\begin{enumerate}
\item[(a)] $\displaystyle\lim_{x\rightarrow2}\dfrac{\sqrt{x+2}-2}{x^{3}-8}$;
\vspace{0.2cm}
Resolu\c{c}\~ao:\\
\begin{align}
&= \dfrac{\sqrt{x+2}-2}{x^3-8} \cdot \dfrac{\sqrt{x+2}+2}{\sqrt{x+2}+2}\\
&=bla\\
\end{align}
\item[(b)] $\displaystyle\lim_{x\rightarrow0^{+}}\left(\dfrac{1}{x}-\dfrac{1}{|x|}\right)$;
\vspace{0.2cm}
\item[(c)] Use o Teorema do confronto para mostrar que $$\lim_{x\rightarrow0}x^{10}\mbox{sen}\left(\dfrac{50\pi}{\sqrt[3]{x}}\right)=0.$$
\end{enumerate}
\end{enumerate}
\end{document}