|
|
| Line 17: |
Line 17: |
| | | | |
| | ===Exemplo 2=== | | ===Exemplo 2=== |
| − | Calcule, caso existam, cada um dos seguintes limites:\\ | + | Calcule, caso existam, cada um dos seguintes limites: |
| − | \begin{tabular}{llll}
| + | '''1.''' $\displaystyle \lim_{x \to + \infty} \frac{3^{x} - 2^{x}}{3^{x+1}+ 2^{x-3}}$ |
| − | \textbf{1.} $\displaystyle \lim_{x \to + \infty} \frac{3^{x} - 2^{x}}{3^{x+1}+ 2^{x-3}}$ & \textbf{2.}
| + | |
| − | $\displaystyle \lim_{x \to +\infty} \frac{3x^2-x-10}{x^2-x-2}$ & \textbf{3.} $\displaystyle \lim_{x \to 2} \frac{\sqrt{x}-\sqrt{2}}{x-2}$ &
| + | |
| − | \textbf{4.} $\displaystyle \lim_{t \to 0} \frac{1-\cos(t)}{\sin(t)}$\\
| + | |
| − | & & & \\
| + | |
| − | \textbf{5.} $\displaystyle \lim_{t \to 2} \frac{e^{2t-4}-1}{t-2}$ & \textbf{6.} $\displaystyle \lim_{t \to +\infty}x^2\sin\frac{1}{x}$ &
| + | |
| − | \textbf{7.} $\displaystyle \lim_{x \to -\infty}xe^x$ &
| + | |
| − | \textbf{8.}$\displaystyle \lim_{x \to +\infty}\left(\sqrt{x}-\sqrt{x+1}\right)$\\
| + | |
| − | & & & \\
| + | |
| − | \textbf{9.} $\displaystyle\lim_{x \to +\infty}\left(\ln(3x^2+2)-\ln (x^2)\right)$ &
| + | |
| − | \textbf{10.} $\displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{\cos(x)}{1-\sin(x)}$ & & \\
| + | |
| − | \end{tabular}\\
| + | |
| | | | |
| − | \textbf{Resolução:}
| + | '''2.''' $\displaystyle \lim_{x \to +\infty} \frac{3x^2-x-10}{x^2-x-2}$ |
| − | \begin{enumerate}
| + | |
| − | \item $\displaystyle \lim_{x \to + \infty} \frac{3^{x} - 2^{x}}{3^{x+1}+ 2^{x-3}}=
| + | |
| − | \displaystyle \lim_{x \to + \infty} \frac{1 - \frac{2^{x}}{3^x}}{3+ \frac{2^{x-3}}{3^x}}=\frac{1}{3}$.\\
| + | |
| − | (Dividimos o numerador e o denominador pela exponencial de maior base, neste caso $3^x$.)\\
| + | |
| | | | |
| − | \item $\displaystyle \lim_{x \to + \infty} \frac{3x^2-x-10}{x^2-x-2}=
| + | '''3.''' $\displaystyle \lim_{x \to 2} \frac{\sqrt{x}-\sqrt{2}}{x-2}$ |
| − | \displaystyle\lim_{x \to + \infty} \frac{3-\frac{1}{x}-\frac{10}{x^2}}{1-\frac{1}{x}-\frac{2}{x^2}}=3$.\\
| + | |
| − | (Dividimos o numerador e o denominador pela maior potência de $x$.)\\
| + | |
| | | | |
| − | \item
| + | '''4.''' $\displaystyle \lim_{t \to 0} \frac{1-\cos(t)}{\sin(t)}$ |
| − | $$
| + | |
| − | \begin{array}{lll}
| + | |
| − | \displaystyle \lim_{x \to 2}{\frac{\sqrt{x}-\sqrt{2}}{x-2}}&
| + | |
| − | = & \displaystyle \lim_{x\to 2}\frac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{(x-2)(\sqrt{x}+\sqrt{2})} \hspace{1cm} \mbox{\small ( multiplicando pelo conjugado)}\\
| + | |
| − | & = &\displaystyle \lim_{x \to 2} \frac{x-2}{(x-2)(\sqrt{x}+\sqrt{2})} \\
| + | |
| − | & = & \displaystyle \lim_{x \to 2}{\frac{1}{\sqrt{x}+\sqrt{2}}} \hspace{1cm} \mbox{\small (como $x \neq 2$ podemos simplificar a expressão)}\\
| + | |
| − | & = & \displaystyle \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}\\
| + | |
| − | \end{array}$$
| + | |
| − | \item $\displaystyle \lim_{t \to 0} \frac{1-\cos(t)}{\sin(t)}=
| + | |
| − | \displaystyle \lim_{t \to 0} \frac{1-\cos(t)}{t}\cdot \frac{t}{\sin(t)}=0\cdot 1=0$.\\
| + | |
| | | | |
| − | \item Como $\displaystyle \lim_{y \to 0} \frac{e^{y}-1}{y}=1$ e $2t-4\to 0$ quando $t\to 2$,
| + | '''5.''' $\displaystyle \lim_{t \to 2} \frac{e^{2t-4}-1}{t-2}$ |
| − | $$\displaystyle \lim_{t \to 2} \frac{e^{2t-4}-1}{t-2}=
| + | |
| − | \displaystyle \lim_{t \to 2}2 \frac{e^{2t-4}-1}{2(t-2)}=2\displaystyle \lim_{t \to 2} \frac{e^{2t-4}-1}{2t-4}=2\cdot 1=2.$$
| + | |
| − | \item Como $\displaystyle \lim_{y \to 0} \frac{\sin\left(y\right)}{y}=1$ e $\displaystyle \lim_{x \to +\infty}\frac{1}{x}=0$, tem-se:
| + | |
| − | $$\displaystyle \lim_{x \to +\infty}x^2\sin\frac{1}{x}=
| + | |
| − | \displaystyle \lim_{x \to +\infty}x\frac{\sin\frac{1}{x}}{\frac{1}{x}}=``+\infty\cdot1''=+\infty .$$
| + | |
| − | \item $\displaystyle \lim_{x \to -\infty}xe^x=\displaystyle \lim_{x \to -\infty}\frac{x}{e^{-x}}
| + | |
| − | =\displaystyle \lim_{y \to +\infty}\frac{-y}{e^{y}}=0$ \hspace{1cm}(Recorremos a uma mudança de variável $y=-x$.)\\
| + | |
| − | \item
| + | |
| − | $$\begin{array}{lll}
| + | |
| − | \displaystyle \lim_{x \to +\infty}\left(\sqrt{x}-\sqrt{x+1}\right)&= &
| + | |
| − | v \lim_{x \to+\infty}\frac{(\sqrt{x}-\sqrt{x+1})(\sqrt{x}+\sqrt{x+1})}
| + | |
| − | {(\sqrt{x}+\sqrt{x+1})} \hspace{1cm} \mbox{\small (multiplicando pelo conjugado.)}\\
| + | |
| − | &= &\displaystyle \lim_{x \to +\infty}\frac{x-(x+1)}{(\sqrt{x}+\sqrt{x+1})}\\
| + | |
| − | & = &
| + | |
| − | \displaystyle \lim_{x \to +\infty}\frac{-1}{(\sqrt{x}+\sqrt{x+1})}=0\\
| + | |
| − | \end{array}$$
| + | |
| | | | |
| | + | '''6.''' $\displaystyle \lim_{t \to +\infty}x^2\sin\frac{1}{x}$ |
| | | | |
| − | \item $\displaystyle\lim_{t \to +\infty}\left(\ln(3x^2+2)-\ln (x^2)\right)=
| + | '''7.''' $\displaystyle \lim_{x \to -\infty}xe^x$ |
| − | \displaystyle \lim_{t \to +\infty}\ln\frac{3x^2+2}{x^2}=\ln\left(\displaystyle \lim_{t \to +\infty}\frac{3x^2+2}{x^2}\right)=\ln3$\\
| + | |
| − | {\small (Usamos as propriedades aritméticas dos logaritmos e a continuidade da função logarítmica, que nos permite ``trocar'' o limite com o logaritmo)}\\
| + | |
| | | | |
| − | \item
| + | '''8.''' $\displaystyle \lim_{x \to +\infty}\left(\sqrt{x}-\sqrt{x+1}\right)$ |
| − | $$\begin{array}{lll}
| + | |
| − | \displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{\cos(x)}{1-\sin(x)}&= & | + | '''9.''' $\displaystyle\lim_{x \to +\infty}\left(\ln(3x^2+2)-\ln (x^2)\right)$ |
| − | \displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{\cos(x)(1+\sin(x))}{(1-\sin(x))(1+\sin(x))}\hspace{1cm} (\mbox{\small repare-se que $x \neq \frac{\pi}{2}$ e portanto $\sin{x} \neq 1$})\\ | + | |
| − | & = &
| + | '''10.''' $\displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{\cos(x)}{1-\sin(x)}$ |
| − | \displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{\cos(x)(1+\sin(x))}{1-\sin^2(x)}\\ | + | |
| − | &= & \displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{\cos(x)(1+\sin(x))}{\cos^2(x)} \hspace{1cm}(\mbox{\small usando a fórmula fundamental da trigonometria})\\
| + | [[Resolução 10|Resolução]] |
| − | & = &
| + | |
| − | \displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{1+\sin(x)}{\cos(x)}=-\infty \hspace{1cm} \left(\mbox{\small quando $\displaystyle x \to \frac{\pi}{2}^+$, tem-se que $\displaystyle \sin{x} \to 1$ e $\displaystyle \cos{x} \to 0^-$ }\right) | + | |
| − | \end{array}$$
| + | |
