What is the limit as x approaches infinity of ln(x)sin(1/x) ?

The limit as x approaches infinity of ln(x)sin(1/x) is 0

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limit as x approaches infinity of ln(x)sin(1/x)

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: x \to \infty lim \frac{- \frac{c o s ( \frac{1}{x} )}{x ^{2}}}{- \frac{1}{x l n ^{2} ( x )}}

= x \to \infty lim \frac{- \frac{c o s ( \frac{1}{x} )}{x ^{2}}}{- \frac{1}{x l n ^{2} ( x )}}

Simplify

\frac{- \frac{c o s ( \frac{1}{x} )}{x ^{2}}}{- \frac{1}{x l n ^{2} ( x )}} : \frac{ln ^{2} ( x ) cos ( \frac{1}{x} )}{x}

= x \to \infty lim (\frac{ln ^{2} ( x ) cos ( \frac{1}{x} )}{x})

Apply L'Hopital's Rule

: x \to \infty lim \frac{\frac{2 l n ( x ) c o s ( \frac{1}{x} )}{x} + \frac{l n ^{2} ( x ) s i n ( \frac{1}{x} )}{x ^{2}}}{1}

= x \to \infty lim \frac{\frac{2 l n ( x ) c o s ( \frac{1}{x} )}{x} + \frac{l n ^{2} ( x ) s i n ( \frac{1}{x} )}{x ^{2}}}{1}

Simplify

\frac{\frac{2 l n ( x ) c o s ( \frac{1}{x} )}{x} + \frac{l n ^{2} ( x ) s i n ( \frac{1}{x} )}{x ^{2}}}{1} : \frac{2 x ln ( x ) cos ( \frac{1}{x} ) + ln ^{2} ( x ) sin ( \frac{1}{x} )}{x ^{2}}

= x \to \infty lim (\frac{2 x ln ( x ) cos ( \frac{1}{x} ) + ln ^{2} ( x ) sin ( \frac{1}{x} )}{x ^{2}})

Apply L'Hopital's Rule

: x \to \infty lim \frac{2 ln ( x ) cos ( \frac{1}{x} ) + 2 cos ( \frac{1}{x} ) + \frac{4 s i n ( \frac{1}{x} ) l n ( x )}{x} - \frac{l n ^{2} ( x ) c o s ( \frac{1}{x} )}{x ^{2}}}{2 x}

= x \to \infty lim \frac{2 ln ( x ) cos ( \frac{1}{x} ) + 2 cos ( \frac{1}{x} ) + \frac{4 s i n ( \frac{1}{x} ) l n ( x )}{x} - \frac{l n ^{2} ( x ) c o s ( \frac{1}{x} )}{x ^{2}}}{2 x}

Simplify

\frac{2 ln ( x ) cos ( \frac{1}{x} ) + 2 cos ( \frac{1}{x} ) + \frac{4 s i n ( \frac{1}{x} ) l n ( x )}{x} - \frac{l n ^{2} ( x ) c o s ( \frac{1}{x} )}{x ^{2}}}{2 x} : \frac{2 x ^{2} ln ( x ) cos ( \frac{1}{x} ) + 2 x ^{2} cos ( \frac{1}{x} ) + 4 x sin ( \frac{1}{x} ) ln ( x ) - ln ^{2} ( x ) cos ( \frac{1}{x} )}{2 x ^{3}}

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

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

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

Apply the Squeeze Theorem:

0

= \frac{1}{2} \cdot 0

Simplify

= 0

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What is the limit as x approaches infinity of ln(x)sin(1/x) ?
The limit as x approaches infinity of ln(x)sin(1/x) is 0