What is the limit as x approaches 0 of 2x^2cos(5/x) ?

The limit as x approaches 0 of 2x^2cos(5/x) is 0

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limit as x approaches 0 of 2x^2cos(5/x)

x \to 0 lim (2 x^{2} cos (\frac{5}{x}))

0

Solution steps

x \to 0 lim (2 x^{2} cos (\frac{5}{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} cos (\frac{5}{x}))

Apply the Squeeze Theorem:

0

= 2 \cdot 0

Simplify

= 0

f (x) = 5 x^{5} ln (x)

\frac{d y}{d x} = \frac{y}{x} + 5 x + 4

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

\frac{\partial}{\partial x} (14 x e^{y})

x^{2} y^{^{'}} = y^{2} + x^{2}

What is the limit as x approaches 0 of 2x^2cos(5/x) ?
The limit as x approaches 0 of 2x^2cos(5/x) is 0