Tratamiento riguroso de los limites
En los problemas del 1 al 14 probar, mediante \(\boldsymbol{\epsilon-\delta}\), el límite indicado.
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\[ \lim\limits_{ x \rightarrow 2 } (4x-3) = 5 \]
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\[ \lim\limits_{ t \rightarrow 4 } (9-3t) = -3 \]
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\[ \lim\limits_{ x \rightarrow -2 } \left( \frac{x}{5} + 1 \right) = \frac{3}{5} \]
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\[ \lim\limits_{ x \rightarrow 2 } x^2 = 4 \]
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\[ \lim\limits_{ x \rightarrow -2 } x^3 = -8 \]
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\[ \lim\limits_{ x \rightarrow 1 } \frac{x^2 - 1}{x-1} = 2 \]
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\[ \lim\limits_{ x \rightarrow 1 } (x^2 + 2x -6 ) = -3 \]
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\[ \lim\limits_{ x \rightarrow -1 } ( 2x^2 + 3x - 4 ) = -5 \]
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\[ \lim\limits_{ x \rightarrow 3 } \frac{4}{x-1} = 2 \]
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\[ \lim\limits_{ x \rightarrow 5 } \frac{6}{4-x} = -6 \]
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\[ \lim\limits_{ x \rightarrow 1 } \frac{2x}{5-2x} = \frac{2}{3} \]
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\[ \lim\limits_{ x \rightarrow 4 } \sqrt{x+5} = 3 \]
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\[ \lim\limits_{ x \rightarrow 0 } \frac{1}{ \mid x \mid + 1 } = 1 \]
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\[ \lim\limits_{ x \rightarrow 1/3} \frac{1+x}{x} = 4 \]
En los problemas del 15 al 19 probar, mediante \(\boldsymbol{\epsilon-\delta}\), el límite indicado.
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\(\lim\limits_{ x \rightarrow a } c = c \), \(c\) es una constante.
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\[ \lim\limits_{ x \rightarrow a } x = a \]
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\(\lim\limits_{ x \rightarrow a } x^n = a^n \)
Sugerencia: \(x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + \ldots + xa^{n-2} + a^{n-1})\).
Sugerencia:
\[ \begin{aligned} x^n - a^n &= (x - a)(x^{n-1} + x^{n-2}a \\[1em] &\hspace{1em} + \ldots + xa^{n-2} + a^{n-1}) \end{aligned} \] -
\(\lim\limits_{ x \rightarrow a } \frac{1}{x} = \frac{1}{a}\)
Sugerencia: seguir esquema del ejemplo 1.2.4 tomando \(\beta = \mid a \mid/2\).
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Probar: \( \lim\limits_{ x \rightarrow a } f(x) = L \Rightarrow \lim\limits_{ x \rightarrow a } \mid f(x) \mid = \mid L \mid\)
Sugerencia: \( \left| \mid f(x) \mid - \mid L \mid \right| \leq \mid f(x) - L \mid\).
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\[ \lim\limits_{ x \rightarrow 0 } x^2 \text{sen} \frac{1}{x} = 0 \]
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\[ \lim\limits_{ x \rightarrow 0 } \frac{ \sqrt{ x^2 + 1 } }{ \mid x \mid + 1 } = 1 \]
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\[ \lim\limits_{ x \rightarrow 0 } x \left[ 2 - \sqrt{2} \cos \left( \frac{1}{x^2} \right) \right] = 0 \]
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\[ \lim\limits_{ x \rightarrow 0 } \frac{ \mid x+1 \mid - \mid x-1 \mid }{ \sqrt{3x^2 + 1} } = 0 \]