En los problemas del 1 al 35, hallar el límite indicado.
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\[ \lim_{x\to 2} \frac{x^2+6}{x^2-3} \]
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\[ \lim_{y\to 0} \left[ \frac{y^2-2y+2}{y-4} + 1 \right] \]
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\[ \lim_{x\to \sqrt{2}} \frac{x^2-2}{x^4+x+1} \]
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\[ \lim_{x\to 1} \sqrt{\frac{2x^2+2}{8x^2+1}} \]
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\[ \lim_{x\to 3} \frac{x^2-9}{x-3} \]
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\[ \lim_{y\to -5} \frac{y^2-25}{y+5} \]
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\[ \lim_{h\to 2} \frac{h-2}{h^2-4} \]
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\[ \lim_{x\to 2} \frac{x^3-8}{x-2} \]
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\[ \lim_{y\to -3} \frac{y^3+27}{y+3} \]
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\[ \lim_{x\to 4} \frac{x^2+4x-32}{x-4} \]
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\[ \lim_{x\to -1} \frac{\frac{1}{2}x^2-\frac{5}{2}x-3}{x+1} \]
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\[ \lim_{x\to -2} \frac{\frac{1}{x+1}+1}{x+2} \]
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\[ \lim_{x\to \frac{1}{2}} \frac{8x^3-1}{6x^2-5x+1} \]
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\[ \lim_{x\to 2} \frac{x^4-16}{x-2} \]
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\[ \lim_{x\to 8} \frac{16-x^{4/3}}{4-x^{2/3}} \]
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\[ \lim_{x\to 2} \frac{\sqrt{x^2+5}-3}{x^2-2x} \]
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\[ \lim_{x\to 9} \frac{x^2-81}{\sqrt{x}-3} \]
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\[ \lim_{x\to 0} \frac{x}{\sqrt{x+2}-\sqrt{2}} \]
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\[ \lim_{y\to 0} \frac{\sqrt{y+3}-\sqrt{3}}{y} \]
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\[ \lim_{x\to 1} \frac{\sqrt{x+3}-2}{x-1} \]
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\[ \lim_{y\to 5} \frac{\sqrt{y-1}-2}{y-5} \]
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\[ \lim_{h\to 0} \frac{\sqrt{1+h^2}-1}{h} \]
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\[ \lim_{x\to 7} \frac{2-\sqrt{x-3}}{x^2-49} \]
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\[ \lim_{x\to 1} \frac{x^2-\sqrt{x}}{\sqrt{x}-1} \]
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\[ \lim_{x\to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}} \]
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\[ \lim_{x\to 8} \frac{x-8}{\sqrt[3]{x}-2} \]
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\[ \lim_{x\to 0} \frac{\sqrt[3]{x^2+1}-1}{x^2} \]
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\[ \lim_{x\to 0} \frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x} \]
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\[ \lim_{x\to 1} \frac{\sqrt{x}-1}{\sqrt[3]{x}-1} \]
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\[ \lim_{x\to 64} \frac{\sqrt{x}-8}{\sqrt[3]{x}-4} \]
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\[ \lim_{x\to 1} \frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1} \]
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\[ \lim_{x\to 1} \frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1} \]
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\[ \lim_{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \]
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\[ \lim_{x\to a} \frac{x^3-a^3}{x^2-ax-x+a} \]
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\[ \lim_{x\to 1} \frac{\sqrt{ax+b}-\sqrt{bx+a}}{\sqrt{cx+d}-\sqrt{dx+c}} \]
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\(\text{Si } g(x)=\frac{1}{x}, \, x \neq 0, \text{ probar que:}\) \[ \lim_{h\to 0} \frac{g(x+h)-g(x)}{h} = -\frac{1}{x^2} \]
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\(\text{Si } f(x)=\sqrt{x}, \, x > 0, \text{ probar que:}\) \[ \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \frac{1}{2\sqrt{x}} \]
En los problemas del 38 al 54, hallar el límite indicado.
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\[ \lim_{x\to 2^+} \frac{\sqrt{x-2}}{2x-1} \]
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\[ \lim_{x\to 4^+} \frac{x-4}{\sqrt{x^2-16}} \]
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\[ \lim_{x\to 2^-} \lfloor x \rfloor \]
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\[ \lim_{x\to 2^+} \lfloor x \rfloor \]
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\[ \lim_{x\to -2^-} \lfloor x \rfloor \]
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\[ \lim_{x\to -2^+} \lfloor x \rfloor \]
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\[ \lim_{x\to 0} x^2 \left\lfloor \frac{1}{x} \right\rfloor \]
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\[ \lim_{x\to 2^-} (x – \lfloor x \rfloor) \]
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\[ \lim_{x\to 2^+} (x – \lfloor x \rfloor) \]
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\[ \lim_{x\to 3^-} \lfloor x^2+x+1 \rfloor \]
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\[ \lim_{x\to 3^+} \lfloor x^2+x+1 \rfloor \]
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\[ \lim_{x\to 3^-} (\lfloor x \rfloor + \lfloor 4-x \rfloor) \]
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\[ \lim_{x\to 3^+} (\lfloor x \rfloor – \lfloor 4-x \rfloor) \]
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\[ \lim_{x\to 4^+} \frac{x-4}{|x+4|} \]
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\[ \lim_{x\to 1^+} \frac{\sqrt{x+4}-\sqrt{4x+1}}{\sqrt{x-1}} \]
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\[ \lim_{x\to 2^-} \frac{\sqrt{4-x^2}+2-x}{\sqrt{4-x^3/2} + \sqrt{2x-x^2}} \]
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\[ \lim_{x\to a} \frac{x\sqrt{x}-a\sqrt{a}}{\sqrt[3]{x}-\sqrt[3]{a}} \]
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\[ h(x) = \begin{cases} 2x+1, & \text{si } x \le 2 \\ x^2+1, & \text{si } x > 2 \end{cases} \]
Hallar:
a. \(\lim_{x\to 2^-} h(x)\) b. \(\lim_{x\to 2^+} h(x)\)
c. \(\lim_{x\to 2} h(x)\)
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\[ f(x) = \begin{cases} x^3, & \text{si } x \le 2 \\ x^2+4, & \text{si } x > 2 \end{cases} \]
Hallar:
a. \(\lim_{x\to 2^-} f(x)\) b. \(\lim_{x\to 2^+} f(x)\)
c. \(\lim_{x\to 2} f(x)\)
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\[ f(x) = \begin{cases} -4, & \text{si } x < -2 \\ \frac{x^2}{2}, & \text{si } -2 \le x < 2 \\ x-1, & \text{si } x \ge 2 \end{cases} \]
Hallar:
a. \(\lim_{x\to -2} f(x)\) b. \(\lim_{x\to 2} f(x)\)
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Hallar una función \(f\) tal que \(\lim_{x\to 0^-} f(x) = 3\) y que no exista \(\lim_{x\to 0^+} f(x)\).
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Pruebe, con un contraejemplo, que las proposiciones a continuación son falsas:
- \( \text{Existe } \lim_{x\to a} [f(x) + g(x)] \Rightarrow \text{Existe } \lim_{x\to a} f(x) \text{ y existe } \lim_{x\to a} g(x) \)
- \( \text{Existe } \lim_{x\to a} [f(x)g(x)] \Rightarrow \text{Existe } \lim_{x\to a} f(x) \text{ y existe } \lim_{x\to a} g(x) \)
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Probar que:
\[ \text{Existe } \lim_{x\to a} \frac{f(x)}{g(x)} \text{ y } \lim_{x\to a} g(x) = 0 \Rightarrow \lim_{x\to a} f(x) = 0 \]
De esta proposición se obtiene:
\[ \lim_{x\to a} f(x) \neq 0 \text{ y } \lim_{x\to a} g(x) = 0 \Rightarrow \text{No existe } \lim_{x\to a} \frac{f(x)}{g(x)} \]