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人気のある微分積分 >

limit as x approaches 0+of (ln(x))/(csc(x))

解

x \to 0 + lim (\frac{ln ( x )}{csc ( x )})

解

0

解答ステップ

x \to 0 + lim (\frac{ln ( x )}{csc ( x )})

基本的な三角関数の公式を使用する:

\frac{1}{csc ( x )} = sin (x)

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

ロピタル向けに書き換える

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

ロピタルの定理を適用する

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

簡素化

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

= 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)

= - 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

不定形式の場合を除く

= - \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

= - \frac{1}{\infty}

無限大の特性を適用する:

\frac{c}{\infty} = 0

= 0

グラフ

人気の例

マクローリン f(x)= 1/((1-x)^2)

ma c l a u r in f (x) = \frac{1}{( 1 - x ) ^{2}}

limit as x approaches 0+of 2(tan(x))^x

x \to 0 + lim (2 (tan (x))^{x})

(dy)/(dx)=((x^2+y^2))/(2xy)

\frac{d y}{d x} = \frac{( x ^{2} + y ^{2} )}{2 x y}

y^{''}-y=(e^x)/(e^x+1)

y^{^{''}} - y = \frac{e ^{x}}{e ^{x} + 1}

derivative of ln((x+1/(sqrt(x-2))))

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