Check the continuity of f given by fx- 4-x2 if x- 0 x-5 if 0 - x - 1 4x2-9 if 1 - x - 2 3x-4 if x- 2- at the points 0- 1 and 2-

For the point x=0 RHL = lim_x → 0 4 x^2 = 4 LHL = lim_x → 0 x 5 = 5 As, LHL ≠ RHL it is not continuous at x=0 For the point x=1 ...

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Continuity in an Interval

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Check the continuity of f given by $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} 4 - x^{2} & i f & x \leq 0 x - 5 & i f & 0 < x \leq 1 4 x^{2} - 9 & i f & 1 < x < 2 3 x + 4 & i f & x \geq 2 \end{matrix}$ at the points $0, 1$ and $2$ .

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}} & i f & x < 0 a & i f & x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} & i f & x > 0 \end{matrix}

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - x^{2}, w h e n x \leq 0 5 x - 4, w h e n 0 < x \leq 1 4 x^{2} - 3 x, w h e n 1 < x < 2 3 x + 4, w h e n x \geq 2 \end{matrix}

x

f (x)

f (x) = x, i f x \leq 1

5, i f x \geq 1

f (x)

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} 4 - x^{2} & i f & x \leq 0 x - 5 & i f & 0 < x \leq 1 4 x^{2} - 9 & i f & 1 < x < 2 3 x + 4 & i f & x >\geq 2 \end{matrix}

f (x) = {\begin{matrix} (x^{2} - 9) / (x^{2} - 2 x - 3) & i f & 0 < x < 5 1.5 & i f & x = 3 \end{matrix}

x \neq 3

3

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