Resolução 12
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| − | '''1.''' $\displaystyle \lim_{x \to 1^{-}} f(x)=-\infty$ e $\displaystyle \lim_{x \to 1^{+}} f(x)=+\infty$, | + | '''1.''' $\displaystyle \lim_{x \to 1^{-}} f(x)=-\infty$ e $\displaystyle \lim_{x \to 1^{+}} f(x)=+\infty$, logo a recta $x=1$ é assíntota vertical ao gráfico de $f$. |
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$m=\displaystyle \lim_{x \to \pm\infty} \frac{f(x)}{x}=\displaystyle \lim_{x \to \pm\infty} \frac{x+1}{x^2-x}=0$ e | $m=\displaystyle \lim_{x \to \pm\infty} \frac{f(x)}{x}=\displaystyle \lim_{x \to \pm\infty} \frac{x+1}{x^2-x}=0$ e | ||
| − | $b=\displaystyle \lim_{x \to\pm\infty} f(x)=1$, logo a recta $y=1$ é assímptota horizontal | + | $b=\displaystyle \lim_{x \to\pm\infty} f(x)=1$, logo a recta $y=1$ é assímptota horizontal bilateral ao gráfico de $f$. |
| − | bilateral ao gráfico de $f$. | + | |
| − | + | '''2.''' $\displaystyle \lim_{x \to 0^{+}} g(x)=+\infty$ e $\displaystyle \lim_{x \to 0^{-}} g(x)=0$, logo $x=0$ é assíntota vertical. | |
| − | é | + | |
| − | + | $m=\displaystyle \lim_{x \to +\infty} \frac{g(x)}{x}= \displaystyle \lim_{x \to \pm\infty} \frac{(x+1)e^{\frac{1}{x}}}{x}= | |
| − | $m=\displaystyle \lim_{x \to +\infty} \frac{g(x)}{x}= | + | \displaystyle \lim_{x \to \pm\infty} \left[\left(1+\frac{1}{x}\right)e^{\frac{1}{x}}\right]=1$ |
| − | \displaystyle \lim_{x \to \pm\infty} \frac{(x+1)e^{\frac{1}{x}}}{x}= | + | |
| − | \displaystyle \lim_{x \to \pm\infty} \left[\left(1+\frac{1}{x}\right)e^{\frac{1}{x}}\right]=1$ | + | |
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$ | $ | ||
\begin{aligned} | \begin{aligned} | ||
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\displaystyle \lim_{x \to \pm\infty}\left[(x+1)(e^{\frac{1}{x}}-1)\right]+1= | \displaystyle \lim_{x \to \pm\infty}\left[(x+1)(e^{\frac{1}{x}}-1)\right]+1= | ||
\displaystyle \lim_{x \to \pm\infty}\left[\frac{x+1}{x}\cdot \frac{e^{\frac{1}{x}}-1}{\frac{1}{x}}\right]+1=1 \cdot 1+1=2\end{aligned} | \displaystyle \lim_{x \to \pm\infty}\left[\frac{x+1}{x}\cdot \frac{e^{\frac{1}{x}}-1}{\frac{1}{x}}\right]+1=1 \cdot 1+1=2\end{aligned} | ||
| − | $ | + | $ |
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| + | A reta $y=x+2$ é assíntota oblíqua bilateral ao gráfico de $g$. | ||
| − | + | [[Exemplos ]] | |
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Revision as of 15:39, 15 November 2012
1. $\displaystyle \lim_{x \to 1^{-}} f(x)=-\infty$ e $\displaystyle \lim_{x \to 1^{+}} f(x)=+\infty$, logo a recta $x=1$ é assíntota vertical ao gráfico de $f$.
$m=\displaystyle \lim_{x \to \pm\infty} \frac{f(x)}{x}=\displaystyle \lim_{x \to \pm\infty} \frac{x+1}{x^2-x}=0$ e $b=\displaystyle \lim_{x \to\pm\infty} f(x)=1$, logo a recta $y=1$ é assímptota horizontal bilateral ao gráfico de $f$.
2. $\displaystyle \lim_{x \to 0^{+}} g(x)=+\infty$ e $\displaystyle \lim_{x \to 0^{-}} g(x)=0$, logo $x=0$ é assíntota vertical.
$m=\displaystyle \lim_{x \to +\infty} \frac{g(x)}{x}= \displaystyle \lim_{x \to \pm\infty} \frac{(x+1)e^{\frac{1}{x}}}{x}= \displaystyle \lim_{x \to \pm\infty} \left[\left(1+\frac{1}{x}\right)e^{\frac{1}{x}}\right]=1$
$ \begin{aligned} b=\displaystyle \lim_{x \to +\infty}\left(g(x)-x\right)&= \displaystyle \lim_{x \to \pm\infty}\left((x+1)e^{\frac{1}{x}}-x\right)= \displaystyle \lim_{x \to \pm\infty}\left((x+1)e^{\frac{1}{x}}-(x+1)+1\right)\\&= \displaystyle \lim_{x \to \pm\infty}\left[(x+1)(e^{\frac{1}{x}}-1)\right]+1= \displaystyle \lim_{x \to \pm\infty}\left[\frac{x+1}{x}\cdot \frac{e^{\frac{1}{x}}-1}{\frac{1}{x}}\right]+1=1 \cdot 1+1=2\end{aligned} $
A reta $y=x+2$ é assíntota oblíqua bilateral ao gráfico de $g$.
