What is the limit as x approaches pi/2-of cos(x) ?

The limit as x approaches pi/2-of cos(x) is 0

limit as x approaches pi/2-of cos(x)

x \to \frac{π}{2} - lim (cos (x))

0

Solution steps

x \to \frac{π}{2} - lim (cos (x))

Plug in the value

x = \frac{π}{2}

= cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

\frac{d y}{d x} = \frac{1}{4} y

d er i v a t i v e f (x) = sin (x^{2} + 2)

\frac{d ^{2}}{d x ^{2}} (e^{2 x^{2}})

im pl i c i t \frac{d y}{d x}, e^{9 x} = sin (x + 8 y)

x \to 8 - lim (\frac{∣ x - 8 ∣}{x - 8})

What is the limit as x approaches pi/2-of cos(x) ?
The limit as x approaches pi/2-of cos(x) is 0