limit as x approaches infinity of (2x+5)/(x^2-7x+3)

解

x \to \infty lim (\frac{2 x + 5}{x ^{2} - 7 x + 3})

0

解答ステップ

x \to \infty lim (\frac{2 x + 5}{x ^{2} - 7 x + 3})

最高の分母べき乗で割る:

\frac{\frac{2}{x} + \frac{5}{x ^{2}}}{1 - \frac{7}{x} + \frac{3}{x ^{2}}}

= x \to \infty lim (\frac{\frac{2}{x} + \frac{5}{x ^{2}}}{1 - \frac{7}{x} + \frac{3}{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} ( \frac{2}{x} + \frac{5}{x ^{2}} )}{lim _{x \to \infty} ( 1 - \frac{7}{x} + \frac{3}{x ^{2}} )}

x \to \infty lim (\frac{2}{x} + \frac{5}{x ^{2}}) = 0

x \to \infty lim (1 - \frac{7}{x} + \frac{3}{x ^{2}}) = 1

= \frac{0}{1}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= 0