What is the limit as x approaches 0-of 2x^2sin(1/x) ?

The limit as x approaches 0-of 2x^2sin(1/x) is 0

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limit as x approaches 0-of 2x^2sin(1/x)

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

0

Solution steps

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

Apply the Squeeze Theorem:

0

= 2 \cdot 0

Simplify

= 0

\frac{\partial}{\partial x} (8 + x^{2} - 2 y^{2})

\int \frac{t ^{2} cos ( 2 t ^{3} )}{csc ( 10 t ^{3} )} d t

x \to \infty lim (x^{2} - 1)

\int x (x + 1) d x

\frac{\partial}{\partial y} (ln (x))

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