What is the limit as t approaches 0+of (2sin(4t))/t ?

The limit as t approaches 0+of (2sin(4t))/t is 8

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limit as t approaches 0+of (2sin(4t))/t

t \to 0 + lim (\frac{2 sin ( 4 t )}{t})

8

Solution steps

t \to 0 + lim (\frac{2 sin ( 4 t )}{t})

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

= 2 \cdot t \to 0 + lim (\frac{sin ( 4 t )}{t})

Apply L'Hopital's Rule

: t \to 0 + lim (\frac{cos ( 4 t ) \cdot 4}{1})

= 2 \cdot t \to 0 + lim (\frac{cos ( 4 t ) \cdot 4}{1})

Plug in the value

t = 0

= 2 \cdot \frac{cos ( 4 \cdot 0 ) \cdot 4}{1}

Simplify

2 \cdot \frac{cos ( 4 \cdot 0 ) \cdot 4}{1} : 8

= 8

x \to 0 + lim ((2 x)^{x})

x \to \frac{π}{2} lim (\frac{1}{tan ( x )})

x \to - 6 lim (\frac{x}{x + 6})

x \to \frac{1}{2} lim (x tan (π x))

x \to - \infty lim (x, - 3)

What is the limit as t approaches 0+of (2sin(4t))/t ?
The limit as t approaches 0+of (2sin(4t))/t is 8