limit as x approaches infinity of (6x^2)/(e^{5x)}

解答

x \to \infty lim (\frac{6 x ^{2}}{e ^{5 x}})

0

求解步骤

x \to \infty lim (\frac{6 x ^{2}}{e ^{5 x}})

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

= 6 \cdot x \to \infty lim (\frac{x ^{2}}{e ^{5 x}})

使用指数法则:

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

\frac{x ^{2}}{e ^{5 x}} = (x^{2}) e^{- 5 x}

= 6 \cdot x \to \infty lim (x^{2} e^{- 5 x})

改写以应用洛必达法则

= 6 \cdot x \to \infty lim (\frac{x ^{2}}{\frac{1}{e ^{- 5 x}}})

使用洛必达法则

= 6 \cdot x \to \infty lim (\frac{2 x}{e ^{5 x} \cdot 5})

使用洛必达法则

= 6 \cdot x \to \infty lim (\frac{2}{25 e ^{5 x}})

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

= 6 \cdot \frac{2}{25} \cdot x \to \infty lim (\frac{1}{e ^{5 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

不定型除外

= 6 \cdot \frac{2}{25} \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( e ^{5 x} )}

x \to \infty lim (1) = 1

x \to \infty lim (e^{5 x}) = \infty

= 6 \cdot \frac{2}{25} \cdot \frac{1}{\infty}

化简

6 \cdot \frac{2}{25} \cdot \frac{1}{\infty} : 0

= 0

\frac{\partial}{\partial x} (ln ∣ x y ∣)

x \to - 2 lim (\frac{x ^{2} + 2}{x - 2})

\int e^{\frac{x}{a}} d x

\int \frac{1}{x x + 9} d x

y^{^{''}} + 10 y^{^{'}} + 25 y = 0