What is the limit as x approaches infinity of (3x^3-4x^2+2)/(x^3+2x+7) ?

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 (3x^3-4x^2+2)/(x^3+2x+7)

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

3

Solution steps

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

Divide by highest denominator power:

\frac{3 - \frac{4}{x} + \frac{2}{x ^{3}}}{1 + \frac{2}{x ^{2}} + \frac{7}{x ^{3}}}

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

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

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

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

= \frac{3}{1}

Simplify

= 3

What is the limit as x approaches infinity of (3x^3-4x^2+2)/(x^3+2x+7) ?
The limit as x approaches infinity of (3x^3-4x^2+2)/(x^3+2x+7) is 3