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人気のある微分積分 >

limit as x approaches 1+of 3/(ln(x))-2/(x-1)

解

x \to 1 + lim (\frac{3}{ln ( x )} - \frac{2}{x - 1})

解

\infty

解答ステップ

x \to 1 + lim (\frac{3}{ln ( x )} - \frac{2}{x - 1})

簡素化

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

= x \to 1 + lim (\frac{3 ( x - 1 ) - 2 ln ( x )}{ln ( x ) ( x - 1 )})

ロピタルの定理を適用する

= x \to 1 + lim (\frac{3 - \frac{2}{x}}{\frac{x - 1}{x} + ln ( x )})

簡素化

\frac{3 - \frac{2}{x}}{\frac{x - 1}{x} + ln ( x )} : \frac{3 x - 2}{x - 1 + x ln ( x )}

= x \to 1 + lim (\frac{3 x - 2}{x - 1 + x ln ( x )})

\frac{3 x - 2}{x - 1 + x ln ( x )} = (3 x - 2) \frac{1}{x - 1 + x ln ( x )}

= x \to 1 + lim ((3 x - 2) \frac{1}{x - 1 + x ln ( x )})

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

不定形式の場合を除く

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

x \to 1 + lim (3 x - 2) = 1

x \to 1 + lim (\frac{1}{x - 1 + x ln ( x )}) = \infty

= 1 \cdot \infty

無限大の特性を適用する:

c \cdot \infty = \infty

= \infty

グラフ

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