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

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

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

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

2

Solution steps

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

Apply L'Hopital's Rule

: t \to 0 lim (\frac{cos ( 2 t ) \cdot 2}{1})

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

Plug in the value

t = 0

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

Simplify

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

= 2

\frac{( s + 2 )}{( s ^{2} + 4 s + 13 )}

t^{2} \frac{d x}{d t} + 3 t x = t^{4} lo g_{10} (t) + 1

\frac{1}{x + x ln ( x )}

f (x) = (4 x^{2} + 3) (4 x - 5)

\frac{d}{d x} (ln (25))

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