limit as x approaches infinity of (4^x+3)/(4^{x+1)}

解

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

\frac{1}{4}

十進法表記

0.25

解答ステップ

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

4^{x + 1}

で割る

: \frac{1}{4} + \frac{3}{4 ^{x + 1}}

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

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

不定形式の場合を除く

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

x \to \infty lim (\frac{1}{4}) = \frac{1}{4}

x \to \infty lim (\frac{3}{4 ^{x + 1}}) = 0

= \frac{1}{4} + 0

簡素化

= \frac{1}{4}