Is the function defined by continuous at x = π ?

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Solution

The given function is,

f( x )= x 2 −sinx+5

At x=π, the function becomes,

f( π )= π 2 −sinπ+5 = π 2 +5

Let, lim x→π f( x )= lim x→π ( x 2 −sinx+5 )

Assume that x=π+h then if x→π, h→0.

lim x→π f( x )= lim x→π ( x 2 −sinx+5 ) = lim h→0 ( ( π+h ) 2 −sin( π+h )+5 ) = π 2 −sinπ+5 = π 2 +5

It can be observed that lim x→π f( x )=f( π ).

Therefore, the function is continuous for x=π.

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Is the function defined by $x^{2}$ - sin x+5 continuous at $x = π$ ?

Is the function defined by continuous at x = $π$ ?

f (x) = x^{2} - sin x + 5

x = π

(\frac{π}{6}, \frac{π}{3})

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{\sqrt{2} cos x - 1}{cot x - 1}, & x \neq \frac{π}{4} k, & x = \frac{π}{4} \end{matrix}

f

x = \frac{π}{4}

f

f (x) = \frac{1 - \sqrt{2} sin x}{π - 4 x}

x \neq \frac{π}{4}

= \frac{1}{4}

x \neq \frac{π}{4}

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