What is the limit as x approaches infinity of xsin((2pi)/x) ?

The limit as x approaches infinity of xsin((2pi)/x) is 2pi

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limit as x approaches infinity of xsin((2pi)/x)

x \to \infty lim (x sin (\frac{2 π}{x}))

2 π

Decimal Notation

6.28318 \dots

Solution steps

x \to \infty lim (x sin (\frac{2 π}{x}))

Rewrite for L'Hopital

= x \to \infty lim (\frac{sin ( \frac{2 π}{x} )}{\frac{1}{x}})

Apply L'Hopital's Rule

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

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

Simplify

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

= x \to \infty lim (2 π cos (\frac{2 π}{x}))

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

= 2 π \cdot x \to \infty lim (cos (\frac{2 π}{x}))

Apply the Limit Chain Rule:

1

= 2 π 1

Simplify

= 2 π

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What is the limit as x approaches infinity of xsin((2pi)/x) ?
The limit as x approaches infinity of xsin((2pi)/x) is 2pi