What is the limit as x approaches 0 of+(xsin(1/x)) ?

The limit as x approaches 0 of+(xsin(1/x)) is 0

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limit as x approaches 0 of+(xsin(1/x))

x \to 0 lim (+ (x sin (\frac{1}{x})))

0

Solution steps

x \to 0 lim (x sin (\frac{1}{x}))

Apply the Squeeze Theorem:

0

= 0

y^{^{'''}} + y^{^{''}} = 3 e^{x} + 4 x^{2}

\frac{d y}{d t} - 2 t y = 9 t^{2} e^{t^{2}}

\frac{d y}{d x} = y sin (x), y (0) = 1

\int e^{x y} d x

f (x) = \frac{1 + 16 x ^{2}}{( 4 x ) ^{3}}

What is the limit as x approaches 0 of+(xsin(1/x)) ?
The limit as x approaches 0 of+(xsin(1/x)) is 0