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人気のある微分積分 >

limit as x approaches infinity of sqrt((8x^2-3x)/(16x^2-1))

解

x \to \infty lim (\frac{8 x ^{2} - 3 x}{16 x ^{2} - 1})

解

\frac{1}{2}

+1

十進法表記

0.70710 \dots

解答ステップ

x \to \infty lim (\frac{8 x ^{2} - 3 x}{16 x ^{2} - 1})

最高の分母べき乗で割る:

\frac{8 - \frac{3}{x}}{16 - \frac{1}{x ^{2}}}

= x \to \infty lim (\frac{8 - \frac{3}{x}}{16 - \frac{1}{x ^{2}}})

x \to a lim [f (x)]^{b} = [x \to a lim f (x)]^{b}

不定形式の場合を除く

= x \to \infty lim (\frac{8 - \frac{3}{x}}{16 - \frac{1}{x ^{2}}})

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 \infty} ( 8 - \frac{3}{x} )}{lim _{x \to \infty} ( 16 - \frac{1}{x ^{2}} )}

x \to \infty lim (8 - \frac{3}{x}) = 8

x \to \infty lim (16 - \frac{1}{x ^{2}}) = 16

= \frac{8}{16}

共通因数を約分する:

8

= \frac{1}{2}

グラフ

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