Calcule, se existir, o limite das fun\u00e7\u00f5es dadas nos pontos indicados:<\/p>\n
1.<\/strong> $x \\to f(x) = {e^{\\sqrt[3]{x}}}$, em $ + \\infty $ e em $ – \\infty $; 2.<\/strong> $x \\to f(x) = {e^{ – {x^2}}}$, em $ + \\infty $ e em $ – \\infty $; 3.<\/strong> $x \\to f(x) = \\frac{{{x^5}}}{{{2^x}}}$, em $ + \\infty $ e em $ – \\infty $; 4.<\/strong> $x \\to f(x) = {x^2}\\,{e^{\\frac{1}{x}}}$, em $ + \\infty $ e em $ – \\infty $; 5.<\/strong> $x \\to f(x) = \\frac{{{e^x} – 1}}{{2x}}$, em $ + \\infty $ e em $0$; 6.<\/strong> $x \\to f(x) = \\frac{{{e^x}}}{{1 – {e^x}}}$, em $ – \\infty $, em $0$ e em $ + \\infty $; 7.<\/strong> $x \\to f(x) = \\frac{{{e^x} – {e^3}}}{{x – 3}}$, em $3$; 8.<\/strong> $x \\to f(x) = \\frac{{\\ln x}}{x}$, em ${0^ + }$; 9.<\/strong> $x \\to f(x) = \\frac{x}{{\\ln x}}$, em ${0^ + }$; 10.<\/strong> $x \\to f(x) = x\\ln x$, em ${0^ + }$; 11.<\/strong> $x \\to f(x) = \\frac{{{x^2} – 1}}{{\\ln {x^2}}}$, em $1$; 12.<\/strong> $x \\to f(x) = \\frac{{{e^x}}}{{\\ln x}}$, em $ + \\infty $.
\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } {e^{\\sqrt[3]{x}}} = {e^{\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\sqrt[3]{x}}} =\u00a0 + \\infty $$
\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } {e^{\\sqrt[3]{x}}} = {e^{\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } \\sqrt[3]{x}}} = 0$$
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\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } {e^{ – {x^2}}} = {e^{\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } ( – {x^2})}} = 0$$
\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } {e^{ – {x^2}}} = {e^{\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } ( – {x^2})}} = 0$$
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\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{x^5}}}{{{2^x}}} = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{x^5}}}{{{{\\left( {{e^{\\ln 2}}} \\right)}^x}}} = \\frac{1}{{{{\\left( {\\ln 2} \\right)}^5}}} \\times \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{{\\left( {x\\ln 2} \\right)}^5}}}{{{e^{x\\ln 2}}}} = \\frac{1}{{{{\\left( {\\ln 2} \\right)}^5}}} \\times \\frac{1}{{\\underbrace {\\mathop {\\lim }\\limits_{y \\to\u00a0 + \\infty } \\frac{{{e^y}}}{{{y^5}}}}_{ + \\infty }}} = 0$$
\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } \\frac{{{x^5}}}{{{2^x}}} = \\mathop {\\lim }\\limits_{y \\to\u00a0 + \\infty } \\frac{{{{\\left( { – y} \\right)}^5}}}{{{2^{ – y}}}} = \\underbrace {\\mathop {\\lim }\\limits_{y \\to\u00a0 + \\infty } {{\\left( { – y} \\right)}^5}}_{ – \\infty } \\times \\underbrace {\\mathop {\\lim }\\limits_{y \\to\u00a0 + \\infty } {2^y}}_{ + \\infty } =\u00a0 – \\infty $$
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\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\left( {{x^2}\\,{e^{\\frac{1}{x}}}} \\right) = \\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } {x^2}}_{ + \\infty } \\times \\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } {e^{\\frac{1}{x}}}}_1 =\u00a0 + \\infty $$
\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } \\left( {{x^2}\\,{e^{\\frac{1}{x}}}} \\right) = \\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } {x^2}}_{ + \\infty } \\times \\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } {e^{\\frac{1}{x}}}}_1 =\u00a0 + \\infty $$
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\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{e^x} – 1}}{{2x}} = \\frac{1}{2} \\times \\left( {\\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{e^x}}}{x}}_{ + \\infty } – \\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{1}{x}}_0} \\right) =\u00a0 + \\infty $$
\n$$\\mathop {\\lim }\\limits_{x \\to 0} f(x) = \\mathop {\\lim }\\limits_{x \\to 0} \\frac{{{e^x} – 1}}{{2x}} = \\frac{1}{2} \\times \\underbrace {\\mathop {\\lim }\\limits_{x \\to 0} \\frac{{{e^x} – 1}}{x}}_1 = \\frac{1}{2}$$
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\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } \\frac{{{e^x}}}{{1 – {e^x}}} = \\frac{{\\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } {e^x}}}{{1 – \\mathop {\\lim }\\limits_{x \\to\u00a0 – \\infty } {e^x}}} = \\frac{0}{{1 – 0}} = 0$$
\n$$\\mathop {\\lim }\\limits_{x \\to {0^ – }} f(x) = \\mathop {\\lim }\\limits_{x \\to {0^ – }} \\frac{{{e^x}}}{{1 – {e^x}}} = \\mathop {\\lim }\\limits_{x \\to {0^ – }} \\left( {\\frac{{{e^x}}}{x} \\times \\frac{x}{{1 – {e^x}}}} \\right) =\u00a0 – \\underbrace {\\mathop {\\lim }\\limits_{x \\to {0^ – }} \\frac{{{e^x}}}{x}}_{ – \\infty } \\times \\frac{1}{{\\underbrace {\\mathop {\\lim }\\limits_{x \\to {0^ – }} \\frac{{{e^x} – 1}}{x}}_1}} =\u00a0 + \\infty $$
\n$$\\mathop {\\lim }\\limits_{x \\to {0^ + }} f(x) = \\mathop {\\lim }\\limits_{x \\to {0^ + }} \\frac{{{e^x}}}{{1 – {e^x}}} = \\mathop {\\lim }\\limits_{x \\to {0^ + }} \\left( {\\frac{{{e^x}}}{x} \\times \\frac{x}{{1 – {e^x}}}} \\right) =\u00a0 – \\underbrace {\\mathop {\\lim }\\limits_{x \\to {0^ + }} \\frac{{{e^x}}}{x}}_{ + \\infty } \\times \\frac{1}{{\\underbrace {\\mathop {\\lim }\\limits_{x \\to {0^ + }} \\frac{{{e^x} – 1}}{x}}_1}} =\u00a0 – \\infty $$
\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{e^x}}}{{1 – {e^x}}} = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{1}{{\\frac{{1 – {e^x}}}{{{e^x}}}}} = \\frac{1}{{\\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\left( {\\frac{1}{{{e^x}}} – 1} \\right)}_{ – 1}}} =\u00a0 – 1$$
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\n$$\\mathop {\\lim }\\limits_{x \\to 3} f(x) = \\mathop {\\lim }\\limits_{x \\to 3} \\frac{{{e^x} – {e^3}}}{{x – 3}} = \\mathop {\\lim }\\limits_{y \\to 0} \\frac{{{e^{y + 3}} – {e^3}}}{y} = \\mathop {\\lim }\\limits_{y \\to 0} \\frac{{{e^3}\\left( {{e^y} – 1} \\right)}}{y} = {e^3} \\times \\underbrace {\\mathop {\\lim }\\limits_{y \\to 0} \\frac{{{e^y} – 1}}{y}}_1 = {e^3}$$
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\n$$\\mathop {\\lim }\\limits_{x \\to {0^ + }} f(x) = \\mathop {\\lim }\\limits_{x \\to {0^ + }} \\frac{{\\ln x}}{x} = \\underbrace {\\mathop {\\lim }\\limits_{x \\to {0^ + }} \\left( {\\ln x} \\right)}_{ – \\infty } \\times \\underbrace {\\mathop {\\lim }\\limits_{x \\to {0^ + }} \\frac{1}{x}}_{ + \\infty } =\u00a0 – \\infty $$
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\n$$\\mathop {\\lim }\\limits_{x \\to {0^ + }} f(x) = \\mathop {\\lim }\\limits_{x \\to {0^ + }} \\frac{x}{{\\ln x}} = \\underbrace {\\mathop {\\lim }\\limits_{x \\to {0^ + }} x}_{{0^ + }} \\times \\underbrace {\\mathop {\\lim }\\limits_{x \\to {0^ + }} \\frac{1}{{\\ln x}}}_{{0^ – }} = 0$$
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\n$$\\mathop {\\lim }\\limits_{x \\to {0^ + }} f(x) = \\mathop {\\lim }\\limits_{x \\to {0^ + }} \\left( {x\\ln x} \\right) = \\mathop {\\lim }\\limits_{y \\to\u00a0 – \\infty } \\left( {{e^y} \\times y} \\right) =\u00a0 – \\mathop {\\lim }\\limits_{y \\to\u00a0 + \\infty } \\left( {{e^{ – y}} \\times y} \\right) =\u00a0 – \\frac{1}{{\\underbrace {\\mathop {\\lim }\\limits_{y \\to\u00a0 + \\infty } \\frac{{{e^y}}}{y}}_{ + \\infty }}} = 0$$
\nFazendo $x = {e^y}$, vem $y = \\ln x$; quando $x \\to {0^ + } \\Rightarrow y \\to\u00a0 – \\infty $.
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\n$$\\mathop {\\lim }\\limits_{x \\to 1} f(x) = \\mathop {\\lim }\\limits_{x \\to 1} \\frac{{{x^2} – 1}}{{\\ln {x^2}}} = \\mathop {\\lim }\\limits_{x \\to 1} \\frac{{{x^2} – 1}}{{2\\ln x}} = \\mathop {\\lim }\\limits_{y \\to 0} \\frac{{{{\\left( {{e^y}} \\right)}^2} – 1}}{{2y}} = \\mathop {\\lim }\\limits_{y \\to 0} \\frac{{{e^{2y}} – 1}}{{2y}} = \\mathop {\\lim }\\limits_{z \\to 0} \\frac{{{e^z} – 1}}{z} = 1$$
\nFazendo $y = \\ln x$, vem $x = {e^y}$; quando $x \\to 1 \\Rightarrow y \\to 0$.
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\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{e^x}}}{{\\ln x}} = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{e^x}}}{x} \\times \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{x}{{\\ln x}} = \\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{e^x}}}{x}}_{ + \\infty } \\times \\frac{1}{{\\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{\\ln x}}{x}}_{{0^ + }}}} =\u00a0 + \\infty $$<\/p>\n