Question

Sketch the graph of f, then identify the values of x₀ for which ${\displaystyle \underset{x\to x_0}{lim}}$ f(x) exists.

f(x) = ${\displaystyle \{\begin{array}{ll}\sin x\text{,}& x<0\\ 1-\cos x\text{,}& 0\le x\le \pi \\ \cos x\text{,}& x>\pi \end{array}}$

Answer 1

f(x) = ${\displaystyle \{\begin{array}{ll}\sin x\text{,}& x<0\\ 1-\cos x\text{,}& 0\le x\le \pi \\ \cos x\text{,}& x>\pi \end{array}}$

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	${\displaystyle -\frac{\pi }{2}}$	0	${\displaystyle \frac{\pi }{2}}$	${\displaystyle \pi }$	${\displaystyle \frac{3\pi }{2}}$	2
f(x)	${\displaystyle \sin (-\frac{\pi }{2})}$	1 – cos 0	${\displaystyle 1-\cos \frac{\pi }{2}}$	1 – cos π	${\displaystyle \cos \left(\frac{3\pi }{2}\right)}$	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.