What is the limit as x approaches infinity of ((4x^3-1)/(3x^2+2))sin(1/x) ?

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limit as x approaches infinity of ((4x^3-1)/(3x^2+2))sin(1/x)

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Apply L'Hopital's Rule

: x \to \infty lim \frac{- \frac{c o s ( \frac{1}{x} ) ( 4 x ^{3} - 1 )}{x ^{2}} + 12 x ^{2} sin ( \frac{1}{x} )}{6 x}

= x \to \infty lim \frac{- \frac{c o s ( \frac{1}{x} ) ( 4 x ^{3} - 1 )}{x ^{2}} + 12 x ^{2} sin ( \frac{1}{x} )}{6 x}

Simplify

\frac{- \frac{c o s ( \frac{1}{x} ) ( 4 x ^{3} - 1 )}{x ^{2}} + 12 x ^{2} sin ( \frac{1}{x} )}{6 x} : \frac{- cos ( \frac{1}{x} ) ( 4 x ^{3} - 1 ) + 12 x ^{4} sin ( \frac{1}{x} )}{6 x ^{3}}

= x \to \infty lim (\frac{- cos ( \frac{1}{x} ) ( 4 x ^{3} - 1 ) + 12 x ^{4} sin ( \frac{1}{x} )}{6 x ^{3}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{6} \cdot x \to \infty lim (\frac{- cos ( \frac{1}{x} ) ( 4 x ^{3} - 1 ) + 12 x ^{4} sin ( \frac{1}{x} )}{x ^{3}})

Expand

\frac{- cos ( \frac{1}{x} ) ( 4 x ^{3} - 1 ) + 12 x ^{4} sin ( \frac{1}{x} )}{x ^{3}} : - 4 cos (\frac{1}{x}) + \frac{cos ( \frac{1}{x} )}{x ^{3}} + 12 x sin (\frac{1}{x})

= \frac{1}{6} \cdot x \to \infty lim (- 4 cos (\frac{1}{x}) + \frac{cos ( \frac{1}{x} )}{x ^{3}} + 12 x sin (\frac{1}{x}))

x \to a lim [f (x) \pm g (x)] = x \to a lim f (x) \pm x \to a lim g (x)

With the exception of indeterminate form

= \frac{1}{6} (- x \to \infty lim (4 cos (\frac{1}{x})) + x \to \infty lim (\frac{cos ( \frac{1}{x} )}{x ^{3}}) + x \to \infty lim (12 x sin (\frac{1}{x})))

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

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

x \to \infty lim (12 x sin (\frac{1}{x})) = 12

= \frac{1}{6} (- 4 + 0 + 12)

Simplify

\frac{1}{6} (- 4 + 0 + 12) : \frac{4}{3}

= \frac{4}{3}

x \to 3 - lim (\frac{x ^{3}}{x - 3})

x \to 0 lim (x ln (x^{4}))

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

x \to 0 lim (\frac{3 x - sin ( x )}{x})

x \to 0 lim (x^{8} cos (\frac{1}{x}))

What is the limit as x approaches infinity of ((4x^3-1)/(3x^2+2))sin(1/x) ?
The limit as x approaches infinity of ((4x^3-1)/(3x^2+2))sin(1/x) is 4/3