limit as x approaches infinity+of (2x+1)/(4x^2+x)

解答

x \to \infty + lim (\frac{2 x + 1}{4 x ^{2} + x})

0

求解步骤

x \to \infty lim (\frac{2 x + 1}{4 x ^{2} + x})

除以分母的最高次幂:

\frac{\frac{2}{x} + \frac{1}{x ^{2}}}{4 + \frac{1}{x}}

= x \to \infty lim (\frac{\frac{2}{x} + \frac{1}{x ^{2}}}{4 + \frac{1}{x}})

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

不定型除外

= \frac{lim _{x \to \infty} ( \frac{2}{x} + \frac{1}{x ^{2}} )}{lim _{x \to \infty} ( 4 + \frac{1}{x} )}

x \to \infty lim (\frac{2}{x} + \frac{1}{x ^{2}}) = 0

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

= \frac{0}{4}

使用法则

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

= 0

\frac{y}{x} \cdot \frac{d y}{d x} = - 4 - \frac{y ^{2}}{x ^{2}}

x \to 3 lim (\frac{1}{x})

x y^{^{'}} - y = \frac{x ^{2}}{y} e^{\frac{y}{x}}

t an g e n t y = x^{3} - 2 x + 2, (3, 23)

y^{^{''}} + 6 y^{^{'}} + 25 y = 0