Calcule os seguintes limites, se existirem:<\/p>\n
1.<\/strong> 2.<\/strong> 3.<\/strong> 4.<\/strong> 5.<\/strong> 6.<\/strong> 7.<\/strong> 8.<\/strong> 9.<\/strong> 10.<\/strong> 11.<\/strong> 12.<\/strong> 13.<\/strong>
\nOra,
\n\\[\\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}\\]
\n
\nOra,
\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\lim }\\limits_{x \\to\u00a0 – 1} \\frac{x}{{{{\\left( {x + 1} \\right)}^2}}}}& = &{\\mathop {\\lim }\\limits_{x \\to\u00a0 – 1} x \\times \\mathop {\\lim }\\limits_{x \\to\u00a0 – 1} \\frac{1}{{{{\\left( {x + 1} \\right)}^2}}}} \\\\ \u00a0 {}& = &{ – 1 \\times \\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 – 1} \\frac{1}{{{{\\left( {x + 1} \\right)}^2}}}}_{ + \\infty }} \\\\ \u00a0 {}& = &{ – \\infty } \\end{array}$$
\n
\nOra,
\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\lim }\\limits_{t \\to\u00a0 – \\infty } \\left( {2{t^3} + {t^2} + 1} \\right)}& = &{\\mathop {\\lim }\\limits_{t \\to\u00a0 – \\infty } \\left[ {{t^3}\\left( {2 + \\frac{1}{t} + \\frac{1}{{{t^2}}}} \\right)} \\right]} \\\\ \u00a0 {}& = &{\\mathop {\\lim }\\limits_{t \\to\u00a0 – \\infty } {t^3} \\times \\mathop {\\lim }\\limits_{t \\to\u00a0 – \\infty } \\left( {2 + \\frac{1}{t} + \\frac{1}{{{t^2}}}} \\right)} \\\\ \u00a0 {}& = &{2 \\times \\underbrace {\\mathop {\\lim }\\limits_{t \\to\u00a0 – \\infty } {t^3}}_{ – \\infty }} \\\\ \u00a0 {}& = &{ – \\infty } \\end{array}$$
\n
\nOra,
\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\lim }\\limits_{m \\to\u00a0 – 1} \\frac{{{m^3} + 1}}{{m + 1}}}& = &{\\mathop {\\lim }\\limits_{m \\to\u00a0 – 1} \\frac{{(m + 1)({m^2} – m + 1)}}{{m + 1}}} \\\\ \u00a0 {}& = &{\\mathop {\\lim }\\limits_{m \\to\u00a0 – 1} ({m^2} – m + 1)} \\\\ \u00a0 {}& = &3 \\end{array}$$\u00a0 \u00a0$$\\begin{array}{*{20}{c}} \u00a0 {}&1&0&0&1 \\\\ \u00a0 { – 1}&{}&{ – 1}&1&{ – 1} \\\\ \\hline \u00a0 {}&1&{ – 1}&1&0 \\end{array}$$
\n
\nOra,
\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\lim }\\limits_{r \\to 2} \\frac{{{r^4} – 16}}{{r – 2}}}& = &{\\mathop {\\lim }\\limits_{r \\to 2} \\frac{{({r^2} – 4)({r^2} + 4)}}{{r – 2}}} \\\\ \u00a0 {}& = &{\\mathop {\\lim }\\limits_{r \\to 2} \\frac{{(r – 2)(r + 2)({r^2} + 4)}}{{r – 2}}} \\\\ \u00a0 {}& = &{\\mathop {\\lim }\\limits_{r \\to 2} (r + 2)({r^2} + 4)} \\\\ \u00a0 {}& = &{32} \\end{array}$$
\n
\nOra,
\n$$\\frac{{\\left| { – 3 + x} \\right|}}{{x – 3}} = \\frac{{\\left| {x – 3} \\right|}}{{x – 3}} = \\left\\{ {\\begin{array}{*{20}{c}} \u00a0 {\\begin{array}{*{20}{l}} \u00a0 {\\frac{{ – x + 3}}{{x – 3}}}& \\Leftarrow &{x – 3 < 0} \\end{array}} \\\\ \u00a0 {\\begin{array}{*{20}{l}} \u00a0 {\\frac{{x – 3}}{{x – 3}}}& \\Leftarrow &{x – 3 > 0} \\end{array}} \\end{array}} \\right. = \\left\\{ {\\begin{array}{*{20}{c}} \u00a0 {\\begin{array}{*{20}{l}} \u00a0 { – 1}& \\Leftarrow &{x < 3} \\end{array}} \\\\ \u00a0 {\\begin{array}{*{20}{l}} \u00a0 1& \\Leftarrow &{x > 3} \\end{array}} \\end{array}} \\right.$$ N\u00e3o existe $\\mathop {\\lim }\\limits_{x \\to 3} \\frac{{\\left| { – 3 + x} \\right|}}{{x – 3}}$, pois $\\mathop {\\lim }\\limits_{x \\to {3^ – }} \\frac{{\\left| { – 3 + x} \\right|}}{{x – 3}} =\u00a0 – 1$ e $\\mathop {\\lim }\\limits_{x \\to {3^ + }} \\frac{{\\left| { – 3 + x} \\right|}}{{x – 3}} = 1$.
\n
\nOra,
\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\lim }\\limits_{s \\to 2} \\left[ {\\left( {1 + \\frac{1}{{s – 2}}} \\right) \\div \\frac{s}{{{s^2} – 4}}} \\right]}& = &{\\mathop {\\lim }\\limits_{s \\to 2} \\left( {\\frac{{s – 2 + 1}}{{s – 2}} \\times \\frac{{{s^2} – 4}}{s}} \\right)} \\\\ \u00a0 {}& = &{\\mathop {\\lim }\\limits_{s \\to 2} \\left( {\\frac{{s – 1}}{{s – 2}} \\times \\frac{{\\left( {s + 2} \\right)\\left( {s – 2} \\right)}}{s}} \\right)} \\\\ \u00a0 {}& = &{\\mathop {\\lim }\\limits_{s \\to 2} \\frac{{\\left( {s – 1} \\right)\\left( {s + 2} \\right)}}{s}} \\\\ \u00a0 {}& = &{\\frac{4}{2}} \\\\ \u00a0 {}& = &2 \\end{array}$$
\n
\nOra,
\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\lim }\\limits_{x \\to 1} \\left[ {\\left( {{x^3} – 3x + 2} \\right) \\times \\frac{{2x}}{{{x^2} – 1}}} \\right]}& = &{\\mathop {\\lim }\\limits_{x \\to 1} \\left[ {{{\\left( {x – 1} \\right)}^2}\\left( {x – 2} \\right) \\times \\frac{{2x}}{{\\left( {x – 1} \\right)\\left( {x + 1} \\right)}}} \\right]} \\\\ \u00a0 {}& = &{\\mathop {\\lim }\\limits_{x \\to 1} \\left[ {\\left( {x – 1} \\right)\\left( {x – 2} \\right) \\times \\frac{{2x}}{{\\left( {x + 1} \\right)}}} \\right]} \\\\ \u00a0 {}& = &0 \\end{array}$$ \u00a0 $$\\begin{array}{*{20}{c}} \u00a0 {}&1&0&{ – 3}&2 \\\\ \u00a0 1&{}&1&1&{ – 2} \\\\ \\hline \u00a0 {}&1&1&{ – 2}&0 \\\\ \u00a0 1&{}&1&2&{} \\\\ \\hline \u00a0 {}&1&2&0&{} \\end{array}$$
\n
\nOra,
\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\left( {\\frac{{{x^3} – x – 1}}{{2{x^3} + 3{x^2}}} + \\frac{1}{{{x^2}}}} \\right)}& = &{\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\left( {\\frac{{1 – \\frac{1}{{{x^2}}} – \\frac{1}{{{x^3}}}}}{{2 + \\frac{3}{x}}} + \\frac{1}{{{x^2}}}} \\right)} \\\\ \u00a0 {}& = &{\\frac{{\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\left( {1 – \\frac{1}{{{x^2}}} – \\frac{1}{{{x^3}}}} \\right)}}{{\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\left( {2 + \\frac{3}{x}} \\right)}} + \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{1}{{{x^2}}}} \\\\ \u00a0 {}& = &{\\frac{1}{2} + 0} \\\\ \u00a0 {}& = &{\\frac{1}{2}} \\end{array}$$
\n
\nOra,
\n$$\\frac{{x + \\left| x \\right|}}{{x – 3\\left| {\\text{x}} \\right|}} = \\left\\{ {\\begin{array}{*{20}{c}} \u00a0 {\\frac{{x + \\left( { – x} \\right)}}{{x – 3\\left( { – x} \\right)}}}& \\Leftarrow &{x < 0} \\\\ \u00a0 {\\frac{{x + x}}{{x – 3x}}}& \\Leftarrow &{x > 0} \\end{array}} \\right. = \\left\\{ {\\begin{array}{*{20}{c}} \u00a0 0& \\Leftarrow &{x < 0} \\\\ \u00a0 { – 1}& \\Leftarrow &{x > 0} \\end{array}} \\right.$$ \u00a0 N\u00e3o 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 $\\mathop {\\lim }\\limits_{x \\to {0^ + }} \\frac{{x + \\left| x \\right|}}{{x – 3\\left| {\\text{x}} \\right|}} =\u00a0 – 1$.
\n
\nOra,
\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\lim }\\limits_{x \\to 0} \\frac{{\\left| x \\right| – 2}}{{{x^2} – 4}}}& = &{\\frac{{\\mathop {\\lim }\\limits_{x \\to 0} \\left( {\\left| x \\right| – 2} \\right)}}{{\\mathop {\\lim }\\limits_{x \\to 0} \\left( {{x^2} – 4} \\right)}}} \\\\ \u00a0 {}& = &{\\frac{{ – 2}}{{ – 4}}} \\\\ \u00a0 {}& = &{\\frac{1}{2}} \\end{array}$$
\n
\nOra,
\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\lim }\\limits_{x \\to 0} \\left( {2x + \\frac{{\\sqrt {{x^2}} }}{{\\left| {\\text{x}} \\right|}}} \\right)}& = &{\\mathop {\\lim }\\limits_{x \\to 0} \\left( {2x + \\frac{{\\left| {\\text{x}} \\right|}}{{\\left| {\\text{x}} \\right|}}} \\right)} \\\\ \u00a0 {}& = &{\\mathop {\\lim }\\limits_{x \\to 0} \\left( {2x + {\\text{1}}} \\right)} \\\\ \u00a0 {}& = &1 \\end{array}$$
\n
\nOra,
\n$$\\begin{array}{*{20}{l}} \u00a0 {\\mathop {\\lim }\\limits_{x \\to {3^ – }} \\left( {\\frac{1}{{x – 3}} – \\frac{5}{{x + 2}}} \\right)}& = &{\\underbrace {\\mathop {\\lim }\\limits_{x \\to {3^ – }} \\left( {\\frac{1}{{x – 3}}} \\right)}_{ – \\infty } – \\underbrace {\\mathop {\\lim }\\limits_{x \\to {3^ – }} \\left( {\\frac{5}{{x + 2}}} \\right)}_1} \\\\ \u00a0 {}& = &{ – \\infty } \\end{array}$$<\/p>\n