Calcule os seguintes limites, se existirem

1.
Ora,
\[\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} – 4x + 3}}{{x + 1}}}& = &{\frac{{\mathop {\lim }\limits_{x \to 3} \left( {{x^2} – 4x + 3} \right)}}{{\mathop {\lim }\limits_{x \to 3} \left( {x + 1} \right)}}}\\{}& = &{\frac{{\mathop {\lim }\limits_{x \to 3} {x^2} + \mathop {\lim }\limits_{x \to 3} \left( { – 4x} \right) + \mathop {\lim }\limits_{x \to 3} 3}}{{\mathop {\lim }\limits_{x \to 3} x + \mathop {\lim }\limits_{x \to 3} 1}}}\\{}& = &{\frac{{9 – 12 + 3}}{{3 + 1}}}\\{}& = &0\end{array}\]

2.
Ora,
$$\begin{array}{*{20}{l}}   {\mathop {\lim }\limits_{x \to  – 1} \frac{x}{{{{\left( {x + 1} \right)}^2}}}}& = &{\mathop {\lim }\limits_{x \to  – 1} x \times \mathop {\lim }\limits_{x \to  – 1} \frac{1}{{{{\left( {x + 1} \right)}^2}}}} \\   {}& = &{ – 1 \times \underbrace {\mathop {\lim }\limits_{x \to  – 1} \frac{1}{{{{\left( {x + 1} \right)}^2}}}}_{ + \infty }} \\   {}& = &{ – \infty } \end{array}$$

3.
Ora,
$$\begin{array}{*{20}{l}}   {\mathop {\lim }\limits_{t \to  – \infty } \left( {2{t^3} + {t^2} + 1} \right)}& = &{\mathop {\lim }\limits_{t \to  – \infty } \left[ {{t^3}\left( {2 + \frac{1}{t} + \frac{1}{{{t^2}}}} \right)} \right]} \\   {}& = &{\mathop {\lim }\limits_{t \to  – \infty } {t^3} \times \mathop {\lim }\limits_{t \to  – \infty } \left( {2 + \frac{1}{t} + \frac{1}{{{t^2}}}} \right)} \\   {}& = &{2 \times \underbrace {\mathop {\lim }\limits_{t \to  – \infty } {t^3}}_{ – \infty }} \\   {}& = &{ – \infty } \end{array}$$

4.
Ora,
$$\begin{array}{*{20}{l}}   {\mathop {\lim }\limits_{m \to  – 1} \frac{{{m^3} + 1}}{{m + 1}}}& = &{\mathop {\lim }\limits_{m \to  – 1} \frac{{(m + 1)({m^2} – m + 1)}}{{m + 1}}} \\   {}& = &{\mathop {\lim }\limits_{m \to  – 1} ({m^2} – m + 1)} \\   {}& = &3 \end{array}$$   $$\begin{array}{*{20}{c}}   {}&1&0&0&1 \\   { – 1}&{}&{ – 1}&1&{ – 1} \\ \hline   {}&1&{ – 1}&1&0 \end{array}$$

5.
Ora,
$$\begin{array}{*{20}{l}}   {\mathop {\lim }\limits_{r \to 2} \frac{{{r^4} – 16}}{{r – 2}}}& = &{\mathop {\lim }\limits_{r \to 2} \frac{{({r^2} – 4)({r^2} + 4)}}{{r – 2}}} \\   {}& = &{\mathop {\lim }\limits_{r \to 2} \frac{{(r – 2)(r + 2)({r^2} + 4)}}{{r – 2}}} \\   {}& = &{\mathop {\lim }\limits_{r \to 2} (r + 2)({r^2} + 4)} \\   {}& = &{32} \end{array}$$

6.
Ora,
$$\frac{{\left| { – 3 + x} \right|}}{{x – 3}} = \frac{{\left| {x – 3} \right|}}{{x – 3}} = \left\{ {\begin{array}{*{20}{c}}   {\begin{array}{*{20}{l}}   {\frac{{ – x + 3}}{{x – 3}}}& \Leftarrow &{x – 3 < 0} \end{array}} \\   {\begin{array}{*{20}{l}}   {\frac{{x – 3}}{{x – 3}}}& \Leftarrow &{x – 3 > 0} \end{array}} \end{array}} \right. = \left\{ {\begin{array}{*{20}{c}}   {\begin{array}{*{20}{l}}   { – 1}& \Leftarrow &{x < 3} \end{array}} \\   {\begin{array}{*{20}{l}}   1& \Leftarrow &{x > 3} \end{array}} \end{array}} \right.$$ Não existe $\mathop {\lim }\limits_{x \to 3} \frac{{\left| { – 3 + x} \right|}}{{x – 3}}$, pois $\mathop {\lim }\limits_{x \to {3^ – }} \frac{{\left| { – 3 + x} \right|}}{{x – 3}} =  – 1$ e $\mathop {\lim }\limits_{x \to {3^ + }} \frac{{\left| { – 3 + x} \right|}}{{x – 3}} = 1$.

7.
Ora,
$$\begin{array}{*{20}{l}}   {\mathop {\lim }\limits_{s \to 2} \left[ {\left( {1 + \frac{1}{{s – 2}}} \right) \div \frac{s}{{{s^2} – 4}}} \right]}& = &{\mathop {\lim }\limits_{s \to 2} \left( {\frac{{s – 2 + 1}}{{s – 2}} \times \frac{{{s^2} – 4}}{s}} \right)} \\   {}& = &{\mathop {\lim }\limits_{s \to 2} \left( {\frac{{s – 1}}{{s – 2}} \times \frac{{\left( {s + 2} \right)\left( {s – 2} \right)}}{s}} \right)} \\   {}& = &{\mathop {\lim }\limits_{s \to 2} \frac{{\left( {s – 1} \right)\left( {s + 2} \right)}}{s}} \\   {}& = &{\frac{4}{2}} \\   {}& = &2 \end{array}$$

8.
Ora,
$$\begin{array}{*{20}{l}}   {\mathop {\lim }\limits_{x \to 1} \left[ {\left( {{x^3} – 3x + 2} \right) \times \frac{{2x}}{{{x^2} – 1}}} \right]}& = &{\mathop {\lim }\limits_{x \to 1} \left[ {{{\left( {x – 1} \right)}^2}\left( {x – 2} \right) \times \frac{{2x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}} \right]} \\   {}& = &{\mathop {\lim }\limits_{x \to 1} \left[ {\left( {x – 1} \right)\left( {x – 2} \right) \times \frac{{2x}}{{\left( {x + 1} \right)}}} \right]} \\   {}& = &0 \end{array}$$   $$\begin{array}{*{20}{c}}   {}&1&0&{ – 3}&2 \\   1&{}&1&1&{ – 2} \\ \hline   {}&1&1&{ – 2}&0 \\   1&{}&1&2&{} \\ \hline   {}&1&2&0&{} \end{array}$$

9.
Ora,
$$\begin{array}{*{20}{l}}   {\mathop {\lim }\limits_{x \to  + \infty } \left( {\frac{{{x^3} – x – 1}}{{2{x^3} + 3{x^2}}} + \frac{1}{{{x^2}}}} \right)}& = &{\mathop {\lim }\limits_{x \to  + \infty } \left( {\frac{{1 – \frac{1}{{{x^2}}} – \frac{1}{{{x^3}}}}}{{2 + \frac{3}{x}}} + \frac{1}{{{x^2}}}} \right)} \\   {}& = &{\frac{{\mathop {\lim }\limits_{x \to  + \infty } \left( {1 – \frac{1}{{{x^2}}} – \frac{1}{{{x^3}}}} \right)}}{{\mathop {\lim }\limits_{x \to  + \infty } \left( {2 + \frac{3}{x}} \right)}} + \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{{x^2}}}} \\   {}& = &{\frac{1}{2} + 0} \\   {}& = &{\frac{1}{2}} \end{array}$$

10.
Ora,
$$\frac{{x + \left| x \right|}}{{x – 3\left| {\text{x}} \right|}} = \left\{ {\begin{array}{*{20}{c}}   {\frac{{x + \left( { – x} \right)}}{{x – 3\left( { – x} \right)}}}& \Leftarrow &{x < 0} \\   {\frac{{x + x}}{{x – 3x}}}& \Leftarrow &{x > 0} \end{array}} \right. = \left\{ {\begin{array}{*{20}{c}}   0& \Leftarrow &{x < 0} \\   { – 1}& \Leftarrow &{x > 0} \end{array}} \right.$$   Não existe $\mathop {\lim }\limits_{x \to 0} \frac{{x + \left| x \right|}}{{x – 3\left| {\text{x}} \right|}}$, pois $\mathop {\lim }\limits_{x \to {0^ – }} \frac{{x + \left| x \right|}}{{x – 3\left| {\text{x}} \right|}} = 0$ e $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{x + \left| x \right|}}{{x – 3\left| {\text{x}} \right|}} =  – 1$.

11.
Ora,
$$\begin{array}{*{20}{l}}   {\mathop {\lim }\limits_{x \to 0} \frac{{\left| x \right| – 2}}{{{x^2} – 4}}}& = &{\frac{{\mathop {\lim }\limits_{x \to 0} \left( {\left| x \right| – 2} \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( {{x^2} – 4} \right)}}} \\   {}& = &{\frac{{ – 2}}{{ – 4}}} \\   {}& = &{\frac{1}{2}} \end{array}$$

12.
Ora,
$$\begin{array}{*{20}{l}}   {\mathop {\lim }\limits_{x \to 0} \left( {2x + \frac{{\sqrt {{x^2}} }}{{\left| {\text{x}} \right|}}} \right)}& = &{\mathop {\lim }\limits_{x \to 0} \left( {2x + \frac{{\left| {\text{x}} \right|}}{{\left| {\text{x}} \right|}}} \right)} \\   {}& = &{\mathop {\lim }\limits_{x \to 0} \left( {2x + {\text{1}}} \right)} \\   {}& = &1 \end{array}$$

13.
Ora,
$$\begin{array}{*{20}{l}}   {\mathop {\lim }\limits_{x \to {3^ – }} \left( {\frac{1}{{x – 3}} – \frac{5}{{x + 2}}} \right)}& = &{\underbrace {\mathop {\lim }\limits_{x \to {3^ – }} \left( {\frac{1}{{x – 3}}} \right)}_{ – \infty } – \underbrace {\mathop {\lim }\limits_{x \to {3^ – }} \left( {\frac{5}{{x + 2}}} \right)}_1} \\   {}& = &{ – \infty } \end{array}$$