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

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

\frac{d}{d x} (\frac{3}{5 + x})

\int \frac{1}{x + 4} d x

y^{^{'}} = \frac{2 cos ( y )}{sin ( y )}

\frac{d}{d x} (\frac{4 x ^{2} - x}{x ^{2} + 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