Calcule, se existir, o limite das funções dadas nos pontos indicados

1. $x \to f(x) = {e^{\sqrt[3]{x}}}$, em $ + \infty $ e em $ – \infty $;
$$\mathop {\lim }\limits_{x \to  + \infty } f(x) = \mathop {\lim }\limits_{x \to  + \infty } {e^{\sqrt[3]{x}}} = {e^{\mathop {\lim }\limits_{x \to  + \infty } \sqrt[3]{x}}} =  + \infty $$
$$\mathop {\lim }\limits_{x \to  – \infty } f(x) = \mathop {\lim }\limits_{x \to  – \infty } {e^{\sqrt[3]{x}}} = {e^{\mathop {\lim }\limits_{x \to  – \infty } \sqrt[3]{x}}} = 0$$

2. $x \to f(x) = {e^{ – {x^2}}}$, em $ + \infty $ e em $ – \infty $;
$$\mathop {\lim }\limits_{x \to  + \infty } f(x) = \mathop {\lim }\limits_{x \to  + \infty } {e^{ – {x^2}}} = {e^{\mathop {\lim }\limits_{x \to  + \infty } ( – {x^2})}} = 0$$
$$\mathop {\lim }\limits_{x \to  – \infty } f(x) = \mathop {\lim }\limits_{x \to  – \infty } {e^{ – {x^2}}} = {e^{\mathop {\lim }\limits_{x \to  – \infty } ( – {x^2})}} = 0$$

3. $x \to f(x) = \frac{{{x^5}}}{{{2^x}}}$, em $ + \infty $ e em $ – \infty $;
$$\mathop {\lim }\limits_{x \to  + \infty } f(x) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^5}}}{{{2^x}}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^5}}}{{{{\left( {{e^{\ln 2}}} \right)}^x}}} = \frac{1}{{{{\left( {\ln 2} \right)}^5}}} \times \mathop {\lim }\limits_{x \to  + \infty } \frac{{{{\left( {x\ln 2} \right)}^5}}}{{{e^{x\ln 2}}}} = \frac{1}{{{{\left( {\ln 2} \right)}^5}}} \times \frac{1}{{\underbrace {\mathop {\lim }\limits_{y \to  + \infty } \frac{{{e^y}}}{{{y^5}}}}_{ + \infty }}} = 0$$
$$\mathop {\lim }\limits_{x \to  – \infty } f(x) = \mathop {\lim }\limits_{x \to  – \infty } \frac{{{x^5}}}{{{2^x}}} = \mathop {\lim }\limits_{y \to  + \infty } \frac{{{{\left( { – y} \right)}^5}}}{{{2^{ – y}}}} = \underbrace {\mathop {\lim }\limits_{y \to  + \infty } {{\left( { – y} \right)}^5}}_{ – \infty } \times \underbrace {\mathop {\lim }\limits_{y \to  + \infty } {2^y}}_{ + \infty } =  – \infty $$

4. $x \to f(x) = {x^2}\,{e^{\frac{1}{x}}}$, em $ + \infty $ e em $ – \infty $;
$$\mathop {\lim }\limits_{x \to  + \infty } f(x) = \mathop {\lim }\limits_{x \to  + \infty } \left( {{x^2}\,{e^{\frac{1}{x}}}} \right) = \underbrace {\mathop {\lim }\limits_{x \to  + \infty } {x^2}}_{ + \infty } \times \underbrace {\mathop {\lim }\limits_{x \to  + \infty } {e^{\frac{1}{x}}}}_1 =  + \infty $$
$$\mathop {\lim }\limits_{x \to  – \infty } f(x) = \mathop {\lim }\limits_{x \to  – \infty } \left( {{x^2}\,{e^{\frac{1}{x}}}} \right) = \underbrace {\mathop {\lim }\limits_{x \to  – \infty } {x^2}}_{ + \infty } \times \underbrace {\mathop {\lim }\limits_{x \to  – \infty } {e^{\frac{1}{x}}}}_1 =  + \infty $$

5. $x \to f(x) = \frac{{{e^x} – 1}}{{2x}}$, em $ + \infty $ e em $0$;
$$\mathop {\lim }\limits_{x \to  + \infty } f(x) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{e^x} – 1}}{{2x}} = \frac{1}{2} \times \left( {\underbrace {\mathop {\lim }\limits_{x \to  + \infty } \frac{{{e^x}}}{x}}_{ + \infty } – \underbrace {\mathop {\lim }\limits_{x \to  + \infty } \frac{1}{x}}_0} \right) =  + \infty $$
$$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} – 1}}{{2x}} = \frac{1}{2} \times \underbrace {\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} – 1}}{x}}_1 = \frac{1}{2}$$

6. $x \to f(x) = \frac{{{e^x}}}{{1 – {e^x}}}$, em $ – \infty $, em $0$ e em $ + \infty $;
$$\mathop {\lim }\limits_{x \to  – \infty } f(x) = \mathop {\lim }\limits_{x \to  – \infty } \frac{{{e^x}}}{{1 – {e^x}}} = \frac{{\mathop {\lim }\limits_{x \to  – \infty } {e^x}}}{{1 – \mathop {\lim }\limits_{x \to  – \infty } {e^x}}} = \frac{0}{{1 – 0}} = 0$$
$$\mathop {\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{x \to {0^ – }} \frac{{{e^x}}}{{1 – {e^x}}} = \mathop {\lim }\limits_{x \to {0^ – }} \left( {\frac{{{e^x}}}{x} \times \frac{x}{{1 – {e^x}}}} \right) =  – \underbrace {\mathop {\lim }\limits_{x \to {0^ – }} \frac{{{e^x}}}{x}}_{ – \infty } \times \frac{1}{{\underbrace {\mathop {\lim }\limits_{x \to {0^ – }} \frac{{{e^x} – 1}}{x}}_1}} =  + \infty $$
$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^x}}}{{1 – {e^x}}} = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{{e^x}}}{x} \times \frac{x}{{1 – {e^x}}}} \right) =  – \underbrace {\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^x}}}{x}}_{ + \infty } \times \frac{1}{{\underbrace {\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^x} – 1}}{x}}_1}} =  – \infty $$
$$\mathop {\lim }\limits_{x \to  + \infty } f(x) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{e^x}}}{{1 – {e^x}}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{\frac{{1 – {e^x}}}{{{e^x}}}}} = \frac{1}{{\underbrace {\mathop {\lim }\limits_{x \to  + \infty } \left( {\frac{1}{{{e^x}}} – 1} \right)}_{ – 1}}} =  – 1$$

7. $x \to f(x) = \frac{{{e^x} – {e^3}}}{{x – 3}}$, em $3$;
$$\mathop {\lim }\limits_{x \to 3} f(x) = \mathop {\lim }\limits_{x \to 3} \frac{{{e^x} – {e^3}}}{{x – 3}} = \mathop {\lim }\limits_{y \to 0} \frac{{{e^{y + 3}} – {e^3}}}{y} = \mathop {\lim }\limits_{y \to 0} \frac{{{e^3}\left( {{e^y} – 1} \right)}}{y} = {e^3} \times \underbrace {\mathop {\lim }\limits_{y \to 0} \frac{{{e^y} – 1}}{y}}_1 = {e^3}$$

8. $x \to f(x) = \frac{{\ln x}}{x}$, em ${0^ + }$;
$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{x} = \underbrace {\mathop {\lim }\limits_{x \to {0^ + }} \left( {\ln x} \right)}_{ – \infty } \times \underbrace {\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{x}}_{ + \infty } =  – \infty $$

9. $x \to f(x) = \frac{x}{{\ln x}}$, em ${0^ + }$;
$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{\ln x}} = \underbrace {\mathop {\lim }\limits_{x \to {0^ + }} x}_{{0^ + }} \times \underbrace {\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{\ln x}}}_{{0^ – }} = 0$$

10. $x \to f(x) = x\ln x$, em ${0^ + }$;
$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {x\ln x} \right) = \mathop {\lim }\limits_{y \to  – \infty } \left( {{e^y} \times y} \right) =  – \mathop {\lim }\limits_{y \to  + \infty } \left( {{e^{ – y}} \times y} \right) =  – \frac{1}{{\underbrace {\mathop {\lim }\limits_{y \to  + \infty } \frac{{{e^y}}}{y}}_{ + \infty }}} = 0$$
Fazendo $x = {e^y}$, vem $y = \ln x$; quando $x \to {0^ + } \Rightarrow y \to  – \infty $.

11. $x \to f(x) = \frac{{{x^2} – 1}}{{\ln {x^2}}}$, em $1$;
$$\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} – 1}}{{\ln {x^2}}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} – 1}}{{2\ln x}} = \mathop {\lim }\limits_{y \to 0} \frac{{{{\left( {{e^y}} \right)}^2} – 1}}{{2y}} = \mathop {\lim }\limits_{y \to 0} \frac{{{e^{2y}} – 1}}{{2y}} = \mathop {\lim }\limits_{z \to 0} \frac{{{e^z} – 1}}{z} = 1$$
Fazendo $y = \ln x$, vem $x = {e^y}$; quando $x \to 1 \Rightarrow y \to 0$.

12. $x \to f(x) = \frac{{{e^x}}}{{\ln x}}$, em $ + \infty $.
$$\mathop {\lim }\limits_{x \to  + \infty } f(x) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{e^x}}}{{\ln x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{e^x}}}{x} \times \mathop {\lim }\limits_{x \to  + \infty } \frac{x}{{\ln x}} = \underbrace {\mathop {\lim }\limits_{x \to  + \infty } \frac{{{e^x}}}{x}}_{ + \infty } \times \frac{1}{{\underbrace {\mathop {\lim }\limits_{x \to  + \infty } \frac{{\ln x}}{x}}_{{0^ + }}}} =  + \infty $$