What is the limit as x approaches pi+of sin(x) ?

The limit as x approaches pi+of sin(x) is 0

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limit as x approaches pi+of sin(x)

x \to π + lim (sin (x))

0

Solution steps

x \to π + lim (sin (x))

Plug in the value

x = π

= sin (π)

Use the following trivial identity:

sin (π) = 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{x ^{2}}{( 1 + 2 x ) ^{3}}

\int - sin (2 t) d t

\int_{1}^{4} 8 t ln (t) d t

y^{^{'}} = - 0.02 (y - 20)

n = 1 \sum \infty 2 - \frac{5}{2 ^{n}}

What is the limit as x approaches pi+of sin(x) ?
The limit as x approaches pi+of sin(x) is 0