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

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

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limit as x approaches 0 of sin^5(2x^2)

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

0

Solution steps

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

Plug in the value

x = 0

= sin^{5} (2 \cdot 0^{2})

Simplify

sin^{5} (2 \cdot 0^{2}) : 0

= 0

\frac{d}{d x} (x^{3} - 3 x + 6)

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

g (t) = \frac{3}{t}

f (x) = \frac{3}{x - 1}

y^{^{''}} + (\frac{2}{x}) y^{^{'}} = 2 x + 4

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