What is the limit as x approaches 0 of 2xsin(1/x) ?

The limit as x approaches 0 of 2xsin(1/x) is 0

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limit as x approaches 0 of 2xsin(1/x)

x \to 0 lim (2 x sin (\frac{1}{x}))

0

Solution steps

x \to 0 lim (2 x sin (\frac{1}{x}))

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

= 2 \cdot x \to 0 lim (x sin (\frac{1}{x}))

Apply the Squeeze Theorem:

0

= 2 \cdot 0

Simplify

= 0

y^{^{'}} = \frac{e ^{y} + 1}{e ^{y}}

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

x^{2} - 7, 6 x

\frac{d}{d x} (3 e^{1 - x^{2}})

f (y) = \frac{1}{y ^{2}}

What is the limit as x approaches 0 of 2xsin(1/x) ?
The limit as x approaches 0 of 2xsin(1/x) is 0