En los problemas del 1 al 43, hallar el límite indicado.
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\[ \lim_{x\to a} \frac{x^3 – ax^2 – a^2x + a^3}{x^2 – a^2} \]
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\[ \lim_{x\to 0} \frac{x – e^x + 1}{x^2} \]
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\[ \lim_{x\to \pi} \frac{\operatorname{sen} x}{x – \pi} \]
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\[ \lim_{x\to \pi} \frac{1 + \cos x}{\tan^2 x} \]
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\[ \lim_{x\to \frac{\pi}{4}} \frac{\sec^2 x – 2\tan x}{1 + \cos 4x} \]
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\[ \lim_{x\to 0^-} \frac{\cot x}{\cot 2x} \]
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\[ \lim_{x\to 0^-} \frac{\frac{\pi}{x}}{\cot\left(\frac{\pi x}{2}\right)} \]
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\[ \lim_{x\to 0} \frac{x \tan^{-1} x}{1 – \cos x} \]
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\[ \lim_{x\to +\infty} \frac{\ln x}{\sqrt[3]{x}} \]
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\[ \lim_{x\to 0} \frac{\ln \operatorname{sen} \pi x}{\ln \operatorname{sen} x} \]
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\[ \lim_{x\to 0} \frac{10^x – 5^x}{x^2} \]
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\[ \lim_{x\to +\infty} \frac{\ln \ln x}{\sqrt{x}} \]
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\[ \lim_{x\to \pi} \frac{(x-\pi)^2}{\operatorname{sen}^2 x} \]
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\[ \lim_{x\to 0} \frac{\tan x – \operatorname{sen} x}{\operatorname{sen}^3 x} \]
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\[ \lim_{x\to 0} \frac{e^x + e^{-x} – x^2 – 2}{\operatorname{sen}^2 x – x^2} \]
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\[ \lim_{x\to 0} \frac{x^2 + 2\cos x – 2}{x^4} \]
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\[ \lim_{x\to \frac{\pi}{4}} \frac{\sec^2 x – 2\tan x}{1 + \cos 4x} \]
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\[ \lim_{x\to 1} \left[ \frac{1}{\ln x} – \frac{x}{\ln x} \right] \]
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\[ \lim_{x\to 1} \left[ \frac{x}{x-1} – \frac{1}{\ln x} \right] \]
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\[ \lim_{x\to 0} \left[ \frac{1}{\operatorname{sen}^2 x} – \frac{1}{x^2} \right] \]
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\[ \lim_{x\to 0} \left[ \frac{1}{x \operatorname{sen} x} – \frac{1}{x^2} \right] \]
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\[ \lim_{x\to 0^+} (1 – \cos x) \cot x \]
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\[ \lim_{x\to \frac{\pi}{4}} (1 – \tan x) \sec 2x \]
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\[ \lim_{x\to 1} (1 – x) \tan \frac{\pi x}{2} \]
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\[ \lim_{x\to a} (x^2 – a^2) \tan \frac{\pi x}{2a} \]
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\[ \lim_{x\to +\infty} x^{\frac{1}{x}} \]
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\[ \lim_{x\to 0^+} x^{\operatorname{sen} x} \]
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\[ \lim_{x\to 1} x^{\frac{1}{1-x}} \]
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\[ \lim_{x\to 0^+} (1 – 2x)^{\frac{1}{x}} \]
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\[ \lim_{x\to 0^+} (1 + x^2)^{\frac{1}{x}} \]
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\[ \lim_{x\to 0^+} (\operatorname{sen} x)^{\operatorname{sen} x} \]
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\[ \lim_{x\to 0} (\operatorname{sen} x)^{x^2} \]
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\[ \lim_{x\to 0^+} (\operatorname{sen} x)^{\tan x} \]
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\[ \lim_{x\to 0^+} (\cot x)^{\frac{1}{\ln x}} \]
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\[ \lim_{x\to 0} \frac{\cosh x – 1}{1 – \cos x} \]
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\[ \lim_{x\to 0} \frac{\tan^{-1} 2x}{\tan^{-1} 3x} \]
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\( \lim_{x\to +\infty} (x – \ln(x^2 – 1)) \)
Sugerencia: \( \ln e^x = x \)
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\[ \lim_{x\to 0} (1 + \operatorname{senh} x)^{\frac{2}{x}} \]
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\[ \lim_{x\to 0^+} \left( \frac{1}{x} – \frac{1}{e^x – 1} \right) \]
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\[ \lim_{x\to +\infty} (e^x – x)^{\frac{1}{x}} \]
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\( \lim_{x\to +\infty} \frac{(\ln x)^n}{x} \)
Sugerencia: \( z = \ln x \)
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\[ \lim_{x\to 0} \frac{\tan^{-1} 3x – 3\tan^{-1} x}{x^3} \]
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\( \lim_{x\to +\infty} \frac{\ln x}{\sqrt{x}} \)
Sugerencia: \( z = \sqrt{x} \)
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Si \( f’ \) es continua, probar: \[ \lim_{h\to 0} \frac{f(x+h) – f(x-h)}{2h} = f'(x) \]
Sugerencia: Usar regla de L’Hôpital derivando respecto a \( h \).
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Si \( f» \) es continua, probar: \[ \lim_{h\to 0} \frac{f(x+h) – 2f(x) + f(x-h)}{h^2} = f»(x) \]
Sugerencia: Usar regla de L’Hôpital derivando 2 veces respecto a \( h \).