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

解答

x \to 0 + lim (6 sin (x) ln (x))

0

求解步骤

x \to 0 + lim (6 sin (x) ln (x))

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

= 6 \cdot x \to 0 + lim (sin (x) ln (x))

改写以应用洛必达法则

= 6 \cdot x \to 0 + lim (\frac{ln ( x )}{\frac{1}{s i n ( x )}})

使用洛必达法则

= 6 \cdot x \to 0 + lim (\frac{\frac{1}{x}}{- cot ( x ) csc ( x )})

化简

\frac{\frac{1}{x}}{- cot ( x ) csc ( x )} : - \frac{1}{x cot ( x ) csc ( x )}

= 6 \cdot x \to 0 + lim (- \frac{1}{x cot ( x ) csc ( x )})

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

= 6 (- x \to 0 + lim (\frac{1}{x cot ( x ) csc ( 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 (- \frac{lim _{x \to 0 +} ( 1 )}{lim _{x \to 0 +} ( x cot ( x ) csc ( x ) )})

x \to 0 + lim (1) = 1

x \to 0 + lim (x cot (x) csc (x)) = \infty

= 6 (- \frac{1}{\infty})

化简

6 (- \frac{1}{\infty}) : 0

= 0

\int \frac{x ^{8} + 1}{x ^{9} + 9 x} d x

l a pl a ce t r an s f or m 4^{t}

im pl i c i t \frac{d y}{d x}, y = \frac{6}{3 x ^{2} + 5}

\frac{d}{d x} (\frac{4}{x + 6})

t an g e n t f (x) = - 4 tan (x) - 1, at x = - \frac{π}{4}