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límite cuando x tiende a infinity de (4x^2)/(e^{4x)}

Solución

x \to \infty lim (\frac{4 x ^{2}}{e ^{4 x}})

0

Pasos de solución

x \to \infty lim (\frac{4 x ^{2}}{e ^{4 x}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 4 \cdot x \to \infty lim (\frac{x ^{2}}{e ^{4 x}})

Aplicar las leyes de los exponentes:

\frac{1}{a ^{b}} = a^{- b}

\frac{x ^{2}}{e ^{4 x}} = (x^{2}) e^{- 4 x}

= 4 \cdot x \to \infty lim (x^{2} e^{- 4 x})

Reescribir para ajustar por L’Hopital

= 4 \cdot x \to \infty lim (\frac{x ^{2}}{\frac{1}{e ^{- 4 x}}})

Aplicar la regla de L'Hopital

: x \to \infty lim (\frac{2 x}{e ^{4 x} \cdot 4})

= 4 \cdot x \to \infty lim (\frac{2 x}{e ^{4 x} \cdot 4})

Aplicar la regla de L'Hopital

: x \to \infty lim (\frac{2}{16 e ^{4 x}})

= 4 \cdot x \to \infty lim (\frac{2}{16 e ^{4 x}})

Eliminar los terminos comunes:

2

= 4 \cdot x \to \infty lim (\frac{1}{8 e ^{4 x}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 4 \cdot \frac{1}{8} \cdot x \to \infty lim (\frac{1}{e ^{4 x}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

con excepción de forma indeterminada

= 4 \cdot \frac{1}{8} \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( e ^{4 x} )}

x \to \infty lim (1) = 1

x \to \infty lim (e^{4 x}) = \infty

= 4 \cdot \frac{1}{8} \cdot \frac{1}{\infty}

Simplificar

4 \cdot \frac{1}{8} \cdot \frac{1}{\infty} : 0

= 0

\frac{\partial}{\partial x} (x^{3} + x^{2} \cdot x^{3} - 2 y^{2})

\frac{d}{d x} (\frac{x - 2}{x ^{2} + 1})

\frac{d}{d x} (- 16)

\frac{d}{d x} (5^{t a n (5 x)})

(- 5, 3), (2, - 3)