What is the limit as x approaches infinity of (4x^3-6x^2)/(7x^{7/2)-5x-3} ?

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as x approaches infinity of (4x^3-6x^2)/(7x^{7/2)-5x-3}

x \to \infty lim (\frac{4 x ^{3} - 6 x ^{2}}{7 x ^{\frac{7}{2}} - 5 x - 3})

0

Solution steps

x \to \infty lim (\frac{4 x ^{3} - 6 x ^{2}}{7 x ^{\frac{7}{2}} - 5 x - 3})

Divide by

x^{\frac{7}{2}} : \frac{\frac{4}{x ^{\frac{1}{2}}} - \frac{6}{x ^{\frac{3}{2}}}}{7 - \frac{5}{x ^{\frac{5}{2}}} - \frac{3}{x ^{\frac{7}{2}}}}

= x \to \infty lim (\frac{\frac{4}{x ^{\frac{1}{2}}} - \frac{6}{x ^{\frac{3}{2}}}}{7 - \frac{5}{x ^{\frac{5}{2}}} - \frac{3}{x ^{\frac{7}{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

With the exception of indeterminate form

= \frac{lim _{x \to \infty} ( \frac{4}{x ^{\frac{1}{2}}} - \frac{6}{x ^{\frac{3}{2}}} )}{lim _{x \to \infty} ( 7 - \frac{5}{x ^{\frac{5}{2}}} - \frac{3}{x ^{\frac{7}{2}}} )}

x \to \infty lim (\frac{4}{x ^{\frac{1}{2}}} - \frac{6}{x ^{\frac{3}{2}}}) = 0

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

= \frac{0}{7}

Apply rule

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

= 0

h \to 0 lim (\frac{- 1}{h})

x \to 1 lim ((x - 1) \cdot ln (x - 1))

x \to - 1 lim (cos (2 π x - π))

x \to \infty lim (x^{2} - 6 x)

x \to π lim (x - π)

What is the limit as x approaches infinity of (4x^3-6x^2)/(7x^{7/2)-5x-3} ?
The limit as x approaches infinity of (4x^3-6x^2)/(7x^{7/2)-5x-3} is 0