Introducción intuitiva a los limites
En los problemas del 1 al 35, hallar el límite indicado.
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\[ \lim\limits_{ x \rightarrow 2 } \frac{x^2 + 6}{x^2 - 3} \]
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\[ \lim\limits_{ y \rightarrow 0 } \left[ \frac{y^2 -2y +2}{y-4} + 1 \right] \]
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\[ \lim\limits_{ x \rightarrow \sqrt{2} } \frac{x^2 - 2}{x^4 + x + 1} \]
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\[ \lim\limits_{ x \rightarrow 1 } \sqrt{ \frac{2x^2 + 2}{8x^2 + 1} } \]
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\[ \lim\limits_{ x \rightarrow 3 } \frac{x^2 -9}{x-3} \]
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\[ \lim\limits_{ y \rightarrow -5 } \frac{y^2 - 25}{y + 5} \]
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\[ \lim\limits_{ h \rightarrow 2 } \frac{h - 2}{h^2 - 4} \]
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\[ \lim\limits_{ x \rightarrow 2 } \frac{x^3 - 8}{x-2} \]
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\[ \lim\limits_{ y \rightarrow -3 } \frac{y^3 + 27}{y + 3} \]
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\[ \lim\limits_{ x \rightarrow 4 } \frac{x^2 + 4x -32}{x - 4} \]
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\[ \lim\limits_{ x \rightarrow -1 } \frac{ \frac{1}{2} x^2 - \frac{5}{2} x -3 }{x+1} \]
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\[ \lim\limits_{ x \rightarrow -2 } \frac{ \frac{1}{x+1} + 1 }{ x+2 } \]
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\[ \lim\limits_{ x \rightarrow {\frac{1}{2}} } \frac{8x^3 - 1}{6x^2 -5x + 1} \]
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\[ \lim\limits_{ x \rightarrow 2 } \frac{ x^4 - 16 }{ x - 2 } \]
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\[ \lim\limits_{ x \rightarrow 8 } \frac{16 - x^{4/3}}{4 - x^{2/3}} \]
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\[ \lim\limits_{ x \rightarrow 2 } \frac{ \sqrt{x^2 + 5} - 3 }{x^2 - 2x} \]
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\[ \lim\limits_{ x \rightarrow 9 } \frac{ x^2 - 81 }{ \sqrt{x} - 3 } \]
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\[ \lim\limits_{ x \rightarrow 0 } \frac{x}{ \sqrt{x+2} - \sqrt{2} } \]
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\[ \lim\limits_{ y \rightarrow 0 } \frac{ \sqrt{y+3} - \sqrt{3} }{ y } \]
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\[ \lim\limits_{ x \rightarrow 1 } \frac{ \sqrt{x+3} - 2 }{ x-1 } \]
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\[ \lim\limits_{ y \rightarrow 5 } \frac{ \sqrt{y-1} - 2 }{ y-5 } \]
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\[ \lim\limits_{ h \rightarrow 0 } \frac{ \sqrt{1+h^2} -1 }{ h } \]
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\[ \lim\limits_{ x \rightarrow 7 } \frac{ 2 - \sqrt{x-3} }{ x^2 - 49 } \]
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\[ \lim\limits_{ x \rightarrow 1 } \frac{ x^2 - \sqrt{x} }{ \sqrt{x} - 1} \]
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\[ \lim\limits_{ x \rightarrow 0 } \frac{ \sqrt{1+x} - \sqrt{1-x}}{ \sqrt[3]{1 + x} - \sqrt[3]{1 - x } } \]
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\[ \lim\limits_{ x \rightarrow 8 } \frac{ x - 8 }{ \sqrt[3]{x} -2 } \]
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\[ \lim\limits_{ x \rightarrow 0 } \frac{ \sqrt[3]{x^2 + 1} -1 }{ x^2 } \]
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\[ \lim\limits_{ x \rightarrow 0 } \frac{ \sqrt[3]{1+x} - \sqrt[3]{1-x} }{ x } \]
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\[ \lim\limits_{ x \rightarrow 1 } \frac{ \sqrt{x} - 1 }{ \sqrt[3]{x} - 1 } \]
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\[ \lim\limits_{ x \rightarrow 64 } \frac{ \sqrt{x} - 8 }{ \sqrt[3]{x} - 4 } \]
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\[ \lim\limits_{ x \rightarrow 1 } \frac{ \sqrt[3]{x} -1 }{ \sqrt[4]{x} -1 } \]
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\[ \lim\limits_{ x \rightarrow 1 } \frac{ \sqrt[n]{x} -1 }{ \sqrt[m]{x} -1 } \]
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\[ \lim\limits_{ x \rightarrow 2 } \frac{ \sqrt{6 - x} - 2 }{ \sqrt{3 - x} - 1 } \]
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\[ \lim\limits_{ x \rightarrow a } \frac{ x^3 - a^3 }{ x^2 - ax - x + a } \]
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\[ \lim\limits_{ x \rightarrow 1 } \frac{ \sqrt{ax + b} - \sqrt{bx + a} }{ \sqrt{cx + d} - \sqrt{dx + c} } \]
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Si \( g(x) = \frac{1}{x},\, x \neq 0\), probar que \( \lim\limits_{ h \rightarrow 0 } \frac{ g(x+h) - g(x) }{h} = - \frac{1}{x^2}\)
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Si \( f(x) = \sqrt{x},\, x > 0\), probar que \( \lim\limits_{ h \rightarrow 0 } \frac{ f(x+h) - f(x) }{h} = \frac{1}{2\sqrt{x}}\)
En los problemas del 38 al 54, hallar el límite indicado.
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\[ \lim\limits_{ x \rightarrow 2^{+} } \frac{ \sqrt{x-2} }{ 2x-1 } \]
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\[ \lim\limits_{ x \rightarrow 4^{+} } \frac{ x - 4 }{ \sqrt{ x^2 - 16 } } \]
- \( \lim\limits_{ x \rightarrow 2^{-} }\) [[\(x\)]]
- \( \lim\limits_{ x \rightarrow 2^{+} }\) [[\(x\)]]
- \(\lim\limits_{ x \rightarrow -2^{-} }\) [[\(x\)]]
- \(\lim\limits_{ x \rightarrow -2^{+} }\) [[\(x\)]]
- \(\lim\limits_{ x \rightarrow 0 } x^2\)[[\(\frac{1}{x}\)]]
- \(\lim\limits_{ x \rightarrow 2^{-} } \left( x -\right.\) [[\(x\)]]\(\left. \right)\)
- \(\lim\limits_{ x \rightarrow 2^{+} } \left( x - \right. \) [[\(x\)]]\( \left. \right)\)
- \(\lim\limits_{ x \rightarrow 3^{-} }\) [[\(x^2+x+1\)]]
- \(\lim\limits_{ x \rightarrow 3^{+} }\) [[\(x^2+x+1\)]]
- \(\lim\limits_{ x \rightarrow 3^{-} } \left[ \right.\) [[\(x\)]] \(+\) [[\(4 - x\)]] \(\left. \right]\)
- \(\lim\limits_{ x \rightarrow 3^{+} } \left[ \right.\) [[\(x\)]] \(-\) [[\(4 - x\)]] \(\left. \right]\)
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\[ \lim\limits_{ x \rightarrow 4^{+} } \frac{ x-4 }{ \mid x+4 \mid } \]
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\[ \lim\limits_{ x \rightarrow 1^{+} } \frac{ \sqrt{x+4} - \sqrt{4x+1} }{ \sqrt{x-1} } \]
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\[ \lim\limits_{ x \rightarrow 2^{-} } \cfrac{ \sqrt{4 - x^2 } + 2 - x }{ \sqrt{ 4 - x^3/2 } + \sqrt{ 2x - x^2 }} \]
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\[ \lim\limits_{ x \rightarrow a } \cfrac{ x\sqrt{x} - a\sqrt{a} }{ \sqrt[3]{x} - \sqrt[3]{a} } \]
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\(\begin{aligned}[t] h(x) = \begin{cases} 2x+1, \text{ si } x \leq 2 \\ x^2 + 1, \text{ si } x > 2 \end{cases} \end{aligned}\) Hallar:
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\[ \lim\limits_{ x \rightarrow 2^{-} } h(x) \]
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\[ \lim\limits_{ x \rightarrow 2^{+} } h(x) \]
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\[ \lim\limits_{ x \rightarrow 2 } h(x) \]
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\(\begin{aligned}[t] f(x) = \begin{cases} x^3,\hspace{1.5em} \text{ si } x \leq 2 \\ x^2 + 4, \text{ si } x > 2 \end{cases} \end{aligned}\) Hallar:
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\[ \lim\limits_{ x \rightarrow 2^{-} } f(x) \]
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\[ \lim\limits_{ x \rightarrow 2^{+} } f(x) \]
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\[ \lim\limits_{ x \rightarrow 2 } f(x) \]
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\(\begin{aligned}[t] f(x) = \begin{cases} -4, \text{ si } x<-2 \\ \frac{x^3}{2}, \text{ si } -2 \leq x < 2 \\ x-1, \text{ si } x \geq 2 \end{cases} \end{aligned}\)   Hallar:
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\[ \lim\limits_{ x \rightarrow -2 } f(x) \]
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\[ \lim\limits_{ x \rightarrow 2 } f(x) \]
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Hallar una función \(f\) tal que \(\lim\limits_{ x \rightarrow 0^{-} } f(x) = 3\) y que no exista \(\lim\limits_{ x \rightarrow 0^{+} } f(x)\).
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Pruebe, con un contraejemplo, que las proposiciones a continuación, son falsas:
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Existe \( \lim\limits_{ x \rightarrow a } \left[ f(x) + g(x) \right] \Rightarrow\) Existe \( \lim\limits_{ x \rightarrow a } f(x) \) y existe \( \lim\limits_{ x \rightarrow a } g(x)\)
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Existe \( \lim\limits_{ x \rightarrow a } \left[ f(x) g(x) \right] \Rightarrow \) Existe \( \lim\limits_{ x \rightarrow a } f(x) \) y existe \( \lim\limits_{ x \rightarrow a } g(x)\)
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Probar que:
\[ \text{Existe } \lim\limits_{ x \rightarrow a } \frac{f(x)}{g(x)}\; \text{ y }\; \lim\limits_{ x \rightarrow a } g(x) = 0 \Rightarrow \lim\limits_{ x \rightarrow a } f(x) = 0 \]\[ \begin{aligned} &\text{Existe } \lim\limits_{ x \rightarrow a } \frac{f(x)}{g(x)} \\[1em] &\hspace{4em}\text{ y }\; \lim\limits_{ x \rightarrow a } g(x) = 0 \\[1em] &\hspace{6em} \Rightarrow \lim\limits_{ x \rightarrow a } f(x) = 0 \end{aligned} \]De esta proposición se obtiene:
\[ \lim_{ x \rightarrow a } f(x) \neq 0 \; \text{ y }\; \lim_{ x \rightarrow a } g(x) = 0 \Rightarrow \text{ No existe } \lim_{ x \rightarrow a } \frac{f(x)}{g(x)} \]\[ \begin{aligned} &\lim_{ x \rightarrow a } f(x) \neq 0 \\[1em] &\hspace{2em}\text{ y }\; \lim_{ x \rightarrow a } g(x) = 0 \\[1em] &\hspace{4em} \Rightarrow \text{ No existe } \lim_{ x \rightarrow a } \frac{f(x)}{g(x)} \end{aligned} \]
Ver Respuestas
Respuestas
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\[ 10 \]
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\[ \frac{1}{2} \]
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\[ 0 \]
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\[ \frac{2}{3} \]
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\[ 6 \]
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\[ -10 \]
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\[ \frac{1}{4} \]
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\[ 12 \]
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\[ 27 \]
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\[ 12 \]
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\[ -\frac{7}{2} \]
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\[ -1 \]
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\[ 18 \]
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\[ 32 \]
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\[ 8 \]
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\[ \frac{1}{3} \]
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\[ 108 \]
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\[ 2\sqrt{2} \]
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\[ \frac{1}{2 \sqrt{3}} \]
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\[ \frac{1}{4} \]
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\[ \frac{1}{4} \]
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\[ 0 \]
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\[ -\frac{1}{56} \]
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\[ 3 \]
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\[ \frac{3}{2} \]
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\[ 12 \]
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\[ \frac{1}{3} \]
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\[ \frac{2}{3} \]
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\[ \frac{3}{2} \]
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\[ 3 \]
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\[ \frac{4}{3} \]
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\[ \frac{m}{n} \]
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\[ \frac{1}{2} \]
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\[ \frac{3 a^2}{a – 1} \]
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\[ \frac{(a – b) \sqrt{c + d}}{(c – d)\sqrt{a + b}} \]
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\[ 0 \]
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\[ 0 \]
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\[ 1 \]
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\[ 2 \]
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\[ -3 \]
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\[ -2 \]
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\[ 0 \]
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\[ 1 \]
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\[ 0 \]
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\[ 12 \]
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\[ 13 \]
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\[ 3 \]
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\[ 3 \]
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\[ 1 \]
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\[ 0 \]
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\[ \frac{\sqrt{2}}{1 + \sqrt{3}} \]
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\[ \frac{9}{2} \sqrt[6]{a^7} \]
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\(5\)
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\(5\)
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\(5\)
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\(8\)
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\(8\)
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\(8\)
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\(-4\)
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no existe
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\[ f(x) = \begin{cases} 3, \hspace{1.8em} \text{ si } x \leq 0 \\[.5em] \text{sen} \frac{\pi}{x}, \text{ si } x > 0 \end{cases} \]
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\(f(x) = \frac{x}{ \mid x \mid }\), \(g(x) = – \frac{x}{ \mid x \mid }\) en \(a = 0\)
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las mismas de (a).
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