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

The limit as t approaches 0 of sin(t) is 0

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limit as t approaches 0 of sin(t)

t \to 0 lim (sin (t))

0

Solution steps

t \to 0 lim (sin (t))

Plug in the value

t = 0

= sin (0)

Use the following trivial identity:

sin (0) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

\frac{d}{d x} (2 x^{\frac{5}{3}} - 5 x^{\frac{4}{5}} + 7 x^{\frac{3}{2}} - 8 x + 17)

\int arcsin^{2} (x) d x

f (x) = \frac{x ^{2}}{x - 11}

x \to \infty lim (\frac{e ^{5 x}}{9 x ^{4} + 31 x + 1})

\frac{d}{d x} (sec^{2} (2 x + 4) \cdot 2)

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