Discuss the continuity of the function at x-2 fx - 1-sin x-2-x2- for x-2 -3- for x - -2

x→π/2lim1 sin x(z2 x)^2 h→ 0 x = (z2 +h) underseth → 0lim1 sin(z2+h)(z2 z2 h)^2 =h→ 0lim1 cos hh^2 =12≠ 3 x→x/2lim f(x) ...

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Continuity in an Interval

Discuss the c...

Question

Discuss the continuity of the function at $x = \frac{π}{2}$
$f (x) = \frac{1 - sin x}{{(\frac{π}{2} - x)}^{2}}$ , for $x \neq \frac{π}{2}$
$= 3$ , for $x = \frac{π}{2}$

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\begin{matrix} f (x) = \frac{1 - sin x}{{(\frac{π}{2} - x)}^{2}}, & f o r & x \neq \frac{π}{2} = 3, & f o r & x = \frac{π}{2} \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

x = \frac{π}{2}

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - sin x}{(π - 2 x)^{2}} & x \neq \frac{π}{2} k & x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

k

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{s i n 2 x}{\sqrt{1 - c o s 2 x}} f o r 0 < x \leq π / 2 \frac{c o s}{π - 2 x} f o r \frac{π}{2} < x < π \end{matrix}

x = π / 2

f (x) = \frac{\sqrt{2} - \sqrt{1 + s i n x}}{c o s x}

x \neq π / 2

x = π / 2

f (π / 2)

x = π

f (x) = \frac{\sqrt{2 + c o s x - 1}}{(π - x^{2})}

x \neq π

f (π)

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