Formas indeterminadas. Regla de L’Hôpital
En los problemas del 1 al 43 hallar el límite indicado.
-
\[ \lim\limits_{ x \rightarrow a} \frac{ x^3 - ax^2 - a^2x + a^3 }{ x^2 - a^2 } \]
-
\[ \lim\limits_{ x \rightarrow 0} \frac{ x - \mathrm{e}^x + 1 }{ x^2 } \]
-
\[ \lim\limits_{ x \rightarrow \pi} \frac{\text{ sen } x}{ x - \pi } \]
-
\[ \lim\limits_{ x \rightarrow \pi} \frac{ 1 + \cos x }{ \tan^2 x } \]
-
\[ \lim\limits_{ x \rightarrow \pi/4} \frac{ \sec^2 x - 2 \tan x }{ 1 + \cos 4x } \]
-
\[ \lim\limits_{ x \rightarrow 0^-} \frac{ \cot x }{ \cot 2x } \]
-
\[ \lim\limits_{ x \rightarrow 0^-} \frac{\pi/x}{ \cot ( \pi x/2 ) } \]
-
\[ \lim\limits_{ x \rightarrow 0} \frac{ x \tan^{-1} x }{ 1 - \cos x } \]
-
\[ \lim\limits_{ x \rightarrow +\infty} \frac{\ln x}{ \sqrt[3]{x} } \]
-
\[ \lim\limits_{ x \rightarrow 0} \frac{\ln \text{ sen } nx}{\ln \text{ sen } x} \]
-
\[ \lim\limits_{ x \rightarrow 0} \frac{10^x - 5^x}{x^2} \]
-
\[ \lim\limits_{ x \rightarrow +\infty} \frac{\ln \ln x}{ \sqrt{x} } \]
-
\[ \lim\limits_{ x \rightarrow \pi} \frac{ (x - \pi)^2 }{ \text{ sen}^2 x } \]
-
\[ \lim\limits_{ x \rightarrow 0} \frac{ \tan x - \text{ sen } x }{ \text{ sen}^3 x } \]
-
\[ \lim\limits_{ x \rightarrow 0} \frac{ \mathrm{e}^x + \mathrm{e}^{-x} - x^2 - 2 } { \text{ sen}^2 x - x^2 } \]
-
\[ \lim\limits_{ x \rightarrow 0} \frac{ x^2 + 2 \cos x - 2 }{x^4} \]
-
\[ \lim\limits_{ x \rightarrow \pi/4 } \frac{ \sec^2 x - 2 \tan x }{ 1 + \cos 4x } \]
-
\[ \lim\limits_{ x \rightarrow 1} \left[ \frac{1}{ \ln x } - \frac{x}{\ln x} \right] \]
-
\[ \lim\limits_{ x \rightarrow 1} \left[ \frac{x}{x-1} - \frac{1}{ \ln x } \right] \]
-
\[ \lim\limits_{ x \rightarrow 0} \left[ \frac{1}{ \text{ sen}^2 x } - \frac{1}{x^2} \right] \]
-
\[ \lim\limits_{ x \rightarrow 0} \left[ \frac{1}{ x \text{ sen } x } - \frac{1}{x^2} \right] \]
-
\[ \lim\limits_{ x \rightarrow 0^+} (1 - \cos x) \cot x \]
-
\[ \lim\limits_{ x \rightarrow \pi/4} (1 - \tan x) \sec 2x \]
-
\[ \lim\limits_{ x \rightarrow 1} (1 - x) \tan \frac{ \pi x }{2} \]
-
\[ \lim\limits_{ x \rightarrow a} \left( x^2 - a^2 \right) \tan \frac{\pi x}{ 2a } \]
-
\[ \lim\limits_{ x \rightarrow +\infty} x^{1/x} \]
-
\[ \lim\limits_{ x \rightarrow 0^+} x^{ \text{sen } x } \]
-
\[ \lim\limits_{ x \rightarrow 1} x^{ 1/(1-x) } \]
-
\[ \lim\limits_{ x \rightarrow 0^+} (1 - 2x)^{ 1/x } \]
-
\[ \lim\limits_{ x \rightarrow 0^+} \left( 1 + x^2 \right)^{1/x} \]
-
\[ \lim\limits_{ x \rightarrow 0^+} ( \text{ sen } x )^{\text{sen } x} \]
-
\[ \lim\limits_{ x \rightarrow 0} ( \text{ sen } x )^{x^2} \]
-
\[ \lim\limits_{ x \rightarrow 0^+} (\text{sen } x)^{\tan x} \]
-
\[ \lim\limits_{ x \rightarrow 0^+} ( \cot x )^{1/\ln x} \]
-
\[ \lim\limits_{ x \rightarrow 0} \frac{ \cosh x - 1 }{ 1 - \cos x } \]
-
\[ \lim\limits_{ x \rightarrow 0} \frac{ \tan^{-1} 2x }{ \tan^{-1} 3x } \]
-
\[ \lim\limits_{ x \rightarrow +\infty} \left( x - \ln \left( x^2 - 1 \right) \right) \]
Sugerencia: \(\ln \mathrm{e}^x = x\)
-
\[ \lim\limits_{ x \rightarrow 0} ( 1 + \text{ senh } x )^{2/x} \]
-
\[ \lim\limits_{ x \rightarrow 0^+} \left( \frac{1}{x} - \frac{1}{ \mathrm{e}^x - 1 } \right) \]
-
\[ \lim\limits_{ x \rightarrow +\infty} \left( \mathrm{e}^x - x \right)^{1/x} \]
-
\[ \lim\limits_{ x \rightarrow +\infty} \frac{ (\ln x)^n }{ x } \]
Sugerencia: \(z = \ln x\)
-
\[ \lim\limits_{ x \rightarrow 0} \frac{ \tan^{-1} 3x - 3\tan^{-1} x }{ x^3 } \]
-
\[ \lim\limits_{ x \rightarrow +\infty} \frac{ \ln x }{ \sqrt[n]{x}} \]
Sugerencia: \(z = \sqrt[n]{x}\)
-
Si \(f'\) es continua, probar:
\[ \lim\limits_{ h \rightarrow 0} \frac{ f(x + h) - f(x - h)}{ 2h } = f'(x) \]Sugerencia: Usar regla de L’Hôpital derivando respecto a \(h\).
-
Si \(f''\) es continua, probar:
\[ \lim\limits_{h \rightarrow 0} \frac{ f(x + h) - 2 f(x) + f(x - h) }{h^2} = f''(x) \]\[ \begin{aligned} &\lim\limits_{h \rightarrow 0} \frac{ f(x + h) - 2 f(x) + f(x - h) }{h^2} \\[.5em] &\hspace{10em} = f''(x) \end{aligned} \]Sugerencia: Usar regla de L’Hôpital derivando 2 veces respecto a \(h\).
Ver Respuestas
Respuestas
-
\[ 0 \]
-
\[ -\frac{1}{2} \]
-
\[ -1 \]
-
\[ \frac{1}{2} \]
-
\[ \frac{1}{2} \]
-
\[ 2 \]
-
\[ \frac{\pi^2}{2} \]
-
\[ 2 \]
-
\[ 0 \]
-
\[ 1 \]
-
\[ \frac{ \ln^2 10 – \ln^2 5 }{2} \]
-
\[ 0 \]
-
\[ 1 \]
-
\[ \frac{1}{2} \]
-
\[ -\frac{1}{4} \]
-
\[ \frac{1}{12} \]
-
\[ \frac{1}{2} \]
-
\[ -1 \]
-
\[ \frac{1}{2} \]
-
\[ \frac{1}{3} \]
-
\[ \frac{1}{6} \]
-
\[ 0 \]
-
\[ 1 \]
-
\[ \frac{2}{\pi} \]
-
\[ – \frac{4 a^2}{\pi} \]
-
\[ 1 \]
-
\[ 1 \]
-
\[ \frac{1}{ \mathrm{e} } \]
-
\[ \mathrm{e}^{-2} \]
-
\[ 1 \]
-
\[ 1 \]
-
\[ 1 \]
-
\[ 1 \]
-
\[ \frac{1}{ \mathrm{e} } \]
-
\[ 1 \]
-
\[ \frac{2}{3} \]
-
\[ +\infty \]
-
\[ \mathrm{e}^2 \]
-
\[ \frac{1}{2} \]
-
\[ \mathrm{e} \]
-
\[ 0 \]
-
\[ -8 \]
-
\[ 0 \]