zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

limit as x approaches 0+of x(ln(x))^6

解答

x \to 0 + lim (x (ln (x))^{6})

解答

0

求解步骤

x \to 0 + lim (x (ln (x))^{6})

改写以应用洛必达法则

= x \to 0 + lim (\frac{( ln ( x ) ) ^{6}}{\frac{1}{x}})

使用洛必达法则

= x \to 0 + lim (\frac{\frac{6 l n ^{5} ( x )}{x}}{- \frac{1}{x ^{2}}})

化简

\frac{\frac{6 l n ^{5} ( x )}{x}}{- \frac{1}{x ^{2}}} : - 6 x ln^{5} (x)

= x \to 0 + lim (- 6 x ln^{5} (x))

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

= - 6 \cdot x \to 0 + lim (x ln^{5} (x))

改写以应用洛必达法则

= - 6 \cdot x \to 0 + lim (\frac{ln ^{5} ( x )}{\frac{1}{x}})

使用洛必达法则

= - 6 \cdot x \to 0 + lim (\frac{\frac{5 l n ^{4} ( x )}{x}}{- \frac{1}{x ^{2}}})

化简

\frac{\frac{5 l n ^{4} ( x )}{x}}{- \frac{1}{x ^{2}}} : - 5 x ln^{4} (x)

= - 6 \cdot x \to 0 + lim (- 5 x ln^{4} (x))

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

= - 6 (- 5 \cdot x \to 0 + lim (x ln^{4} (x)))

改写以应用洛必达法则

= - 6 (- 5 \cdot x \to 0 + lim (\frac{ln ^{4} ( x )}{\frac{1}{x}}))

使用洛必达法则

= - 6 (- 5 \cdot x \to 0 + lim (\frac{\frac{4 l n ^{3} ( x )}{x}}{- \frac{1}{x ^{2}}}))

化简

\frac{\frac{4 l n ^{3} ( x )}{x}}{- \frac{1}{x ^{2}}} : - 4 x ln^{3} (x)

= - 6 (- 5 \cdot x \to 0 + lim (- 4 x ln^{3} (x)))

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

= - 6 (- 5 (- 4 \cdot x \to 0 + lim (x ln^{3} (x))))

改写以应用洛必达法则

= - 6 (- 5 (- 4 \cdot x \to 0 + lim (\frac{ln ^{3} ( x )}{\frac{1}{x}})))

使用洛必达法则

= - 6 (- 5 (- 4 \cdot x \to 0 + lim (\frac{\frac{3 l n ^{2} ( x )}{x}}{- \frac{1}{x ^{2}}})))

化简

\frac{\frac{3 l n ^{2} ( x )}{x}}{- \frac{1}{x ^{2}}} : - 3 x ln^{2} (x)

= - 6 (- 5 (- 4 \cdot x \to 0 + lim (- 3 x ln^{2} (x))))

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

= - 6 (- 5 (- 4 (- 3 \cdot x \to 0 + lim (x ln^{2} (x)))))

改写以应用洛必达法则

= - 6 (- 5 (- 4 (- 3 \cdot x \to 0 + lim (\frac{ln ^{2} ( x )}{\frac{1}{x}}))))

使用洛必达法则

= - 6 (- 5 (- 4 (- 3 \cdot x \to 0 + lim (\frac{\frac{2 l n ( x )}{x}}{- \frac{1}{x ^{2}}}))))

化简

\frac{\frac{2 l n ( x )}{x}}{- \frac{1}{x ^{2}}} : - 2 x ln (x)

= - 6 (- 5 (- 4 (- 3 \cdot x \to 0 + lim (- 2 x ln (x)))))

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

= - 6 (- 5 (- 4 (- 3 (- 2 \cdot x \to 0 + lim (x ln (x))))))

改写以应用洛必达法则

= - 6 (- 5 (- 4 (- 3 (- 2 \cdot x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x}})))))

使用洛必达法则

= - 6 (- 5 (- 4 (- 3 (- 2 \cdot x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}})))))

化简

\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}} : - x

= - 6 (- 5 (- 4 (- 3 (- 2 \cdot x \to 0 + lim (- x)))))

代入值

x = 0

= - 6 (- 5 (- 4 (- 3 (- 2 (- 0)))))

化简

= 0

作图

流行的例子

limit as x approaches 5 of (x^2-2)/(x-5)

x \to 5 lim (\frac{x ^{2} - 2}{x - 5})

limit as x approaches-2 of 1/((x+2))

x \to - 2 lim (\frac{1}{( x + 2 )})

limit as x approaches-3 of 1/((x+3)^4)

x \to - 3 lim (\frac{1}{( x + 3 ) ^{4}})

limit as x approaches 0 of (1-x^2)/x

x \to 0 lim (\frac{1 - x ^{2}}{x})

limit as x approaches 0 of e^{1/x}*x

x \to 0 lim (e^{\frac{1}{x}} \cdot x)