Question

Sketch the graph of f, then identify the values of x₀ for which `lim_(x -> x_0)` f(x) exists.

f(x) = `{{:(sin x",", x < 0),(1 - cos x",", 0 ≤ x ≤ pi),(cos x",", x > pi):}`

Answer 1

f(x) = `{{:(sin x",", x < 0),(1 - cos x",", 0 ≤ x ≤ pi),(cos x",", x > pi):}`

From the figure when x = π, y = f(π) = 2.

The function is not defined at x = π since sin x lies in the interval [– 1, 1]

∴ The given function has limits at all points except at x = π

x	`- pi/2`	0	`pi/2`	`pi`	`(3pi)/2`	2
f(x)	`sin (- pi/2)`	1 – cos 0	`1 - cos  pi/2`	1 – cos π	`cos((3pi)/2)`	cos 2π
f(x)	– 1	0	1	1 – (– 1) = 2	0	1

(π, 2) point is not possible since the range of the curve is [– 1, 1] .

Except x₀ = π, the curve has limits.