limit as n approaches infinity of sqrt(n^4+2n^3)-(n^2+n)

解

n \to \infty lim (n^{4} + 2 n^{3} - (n^{2} + n))

- \frac{1}{2}

十進法表記

- 0.5

解答ステップ

n \to \infty lim (n^{4} + 2 n^{3} - (n^{2} + n))

n^{4} + 2 n^{3} - (n^{2} + n)

の共役で乗じる

\frac{- n ^{2}}{n ^{4} + 2 n ^{3} + ( n ^{2} + n )}

= n \to \infty lim (\frac{- n ^{2}}{n ^{4} + 2 n ^{3} + n ^{2} + n})

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

= - n \to \infty lim (\frac{n ^{2}}{n ^{4} + 2 n ^{3} + n ^{2} + n})

最高の分母べき乗で割る:

\frac{1}{1 + \frac{2}{n} + 1 + \frac{1}{n}}

= - n \to \infty lim \frac{1}{1 + \frac{2}{n} + 1 + \frac{1}{n}}

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 _{n \to \infty} ( 1 )}{lim _{n \to \infty} ( 1 + \frac{2}{n} + 1 + \frac{1}{n} )}

n \to \infty lim (1) = 1

n \to \infty lim (1 + \frac{2}{n} + 1 + \frac{1}{n}) = 2

= - \frac{1}{2}