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प्रश्न
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):}`
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उत्तर
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.
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