What is the limit as x approaches infinity of (x+3)/(2x^3-5) ?

The limit as x approaches infinity of (x+3)/(2x^3-5) is 0

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

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

0

Solution steps

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

Divide by highest denominator power:

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

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

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

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

= \frac{0}{2}

Apply rule

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

= 0

3^{x} ln (3)

\int_{1}^{2} 4 r^{2} ln (r) d r

\frac{d}{d x} (ln (\frac{x - 2}{x + 1}))

\int (2 - x)^{2} ln (4 x) d x

What is the limit as x approaches infinity of (x+3)/(2x^3-5) ?
The limit as x approaches infinity of (x+3)/(2x^3-5) is 0