What is the limit as x approaches infinity of (3x-2)/(6x+1) ?

The limit as x approaches infinity of (3x-2)/(6x+1) is 1/2

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limit as x approaches infinity of (3x-2)/(6x+1)

x \to \infty lim (\frac{3 x - 2}{6 x + 1})

\frac{1}{2}

Decimal

0.5

Solution steps

x \to \infty lim (\frac{3 x - 2}{6 x + 1})

Divide by highest denominator power:

\frac{3 - \frac{2}{x}}{6 + \frac{1}{x}}

= x \to \infty lim (\frac{3 - \frac{2}{x}}{6 + \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

With the exception of indeterminate form

= \frac{lim _{x \to \infty} ( 3 - \frac{2}{x} )}{lim _{x \to \infty} ( 6 + \frac{1}{x} )}

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

x \to \infty lim (6 + \frac{1}{x}) = 6

= \frac{3}{6}

Cancel the common factor:

3

= \frac{1}{2}

\frac{d ^{2} y}{d t ^{2}}

f (x) = arccot (e^{x})

y = x^{2} + x^{- 2} + 7 x - 1

\int x^{3} e^{3 x^{4}} d x

\frac{d}{d x} (tan (3 x - 4 y))

What is the limit as x approaches infinity of (3x-2)/(6x+1) ?
The limit as x approaches infinity of (3x-2)/(6x+1) is 1/2