Exemplos 7

Exemplo 1

Calcule, caso existam, cada um dos seguintes limites:

1. $\displaystyle \lim_{x \to +\infty} \frac{1}{x-3}$

2. $\displaystyle \lim_{x \to 3} \frac{1}{x-3}$

3. $\displaystyle \lim_{x \to 2^-} \frac{1}{2-x}$

4. $\displaystyle \lim_{x \to -1^+} \frac{1}{x^2-1}$

5. $\displaystyle \lim_{x \to 2} \ln(x-2)$

6. $\displaystyle \lim_{x \to 0} e^{\frac{1}{x}}$

Resolução

Exemplo 2

Calcule, caso existam, cada um dos seguintes limites:\\ \begin{tabular}{llll} \textbf{1.} UNIQ5c414ae0535ddfc1-MathJax-12-QINU & \textbf{2.} UNIQ5c414ae0535ddfc1-MathJax-13-QINU & \textbf{3.} UNIQ5c414ae0535ddfc1-MathJax-14-QINU & \textbf{4.} UNIQ5c414ae0535ddfc1-MathJax-15-QINU\\ & & & \\ \textbf{5.} UNIQ5c414ae0535ddfc1-MathJax-16-QINU & \textbf{6.} UNIQ5c414ae0535ddfc1-MathJax-17-QINU & \textbf{7.} UNIQ5c414ae0535ddfc1-MathJax-18-QINU & \textbf{8.}UNIQ5c414ae0535ddfc1-MathJax-19-QINU\\ & & & \\ \textbf{9.} UNIQ5c414ae0535ddfc1-MathJax-20-QINU & \textbf{10.} UNIQ5c414ae0535ddfc1-MathJax-21-QINU & & \\ \end{tabular}\\

\textbf{Resolução:} \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 $\displaystyle \lim_{x \to + \infty} \frac{3x^2-x-10}{x^2-x-2}= \displaystyle\lim_{x \to + \infty} \frac{3-\frac{1}{x}-\frac{10}{x^2}}{1-\frac{1}{x}-\frac{2}{x^2}}=3$.\\

   (Dividimos o numerador e o denominador pela maior potência de $x$.)\\

\item $$ \begin{array}{lll} \displaystyle \lim_{x \to 2}{\frac{\sqrt{x}-\sqrt{2}}{x-2}}& = & \displaystyle \lim_{x\to 2}\frac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{(x-2)(\sqrt{x}+\sqrt{2})} \hspace{1cm} \mbox{\small ( multiplicando pelo conjugado)}\\ & = &\displaystyle \lim_{x \to 2} \frac{x-2}{(x-2)(\sqrt{x}+\sqrt{2})} \\ & = & \displaystyle \lim_{x \to 2}{\frac{1}{\sqrt{x}+\sqrt{2}}} \hspace{1cm} \mbox{\small (como $x \neq 2$ podemos simplificar a expressão)}\\ & = & \displaystyle \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}\\ \end{array}$$ \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 $t\to 2$, $$\displaystyle \lim_{t \to 2} \frac{e^{2t-4}-1}{t-2}= \displaystyle \lim_{t \to 2}2 \frac{e^{2t-4}-1}{2(t-2)}=2\displaystyle \lim_{t \to 2} \frac{e^{2t-4}-1}{2t-4}=2\cdot 1=2.$$ \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 $$\begin{array}{lll} \displaystyle \lim_{x \to +\infty}\left(\sqrt{x}-\sqrt{x+1}\right)&= & v \lim_{x \to+\infty}\frac{(\sqrt{x}-\sqrt{x+1})(\sqrt{x}+\sqrt{x+1})} {(\sqrt{x}+\sqrt{x+1})} \hspace{1cm} \mbox{\small (multiplicando pelo conjugado.)}\\ &= &\displaystyle \lim_{x \to +\infty}\frac{x-(x+1)}{(\sqrt{x}+\sqrt{x+1})}\\ & = & \displaystyle \lim_{x \to +\infty}\frac{-1}{(\sqrt{x}+\sqrt{x+1})}=0\\ \end{array}$$

\item $\displaystyle\lim_{t \to +\infty}\left(\ln(3x^2+2)-\ln (x^2)\right)= \displaystyle \lim_{t \to +\infty}\ln\frac{3x^2+2}{x^2}=\ln\left(\displaystyle \lim_{t \to +\infty}\frac{3x^2+2}{x^2}\right)=\ln3$\\ {\small (Usamos as propriedades aritméticas dos logaritmos e a continuidade da função logarítmica, que nos permite ``trocar o limite com o logaritmo)}\\

\item $$\begin{array}{lll} \displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{\cos(x)}{1-\sin(x)}&= & \displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{\cos(x)(1+\sin(x))}{(1-\sin(x))(1+\sin(x))}\hspace{1cm} (\mbox{\small repare-se que $x \neq \frac{\pi}{2}$ e portanto $\sin{x} \neq 1$})\\ & = & \displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{\cos(x)(1+\sin(x))}{1-\sin^2(x)}\\ &= & \displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{\cos(x)(1+\sin(x))}{\cos^2(x)} \hspace{1cm}(\mbox{\small usando a fórmula fundamental da trigonometria})\\ & = & \displaystyle \lim_{x \to \frac{\pi}{2}^+}\frac{1+\sin(x)}{\cos(x)}=-\infty \hspace{1cm} \left(\mbox{\small quando $\displaystyle x \to \frac{\pi}{2}^+$, tem-se que $\displaystyle \sin{x} \to 1$ e $\displaystyle \cos{x} \to 0^-$ }\right) \end{array}$$