# Is the function defined by f(x) = x^{2 }− sin x + 5 continuous at x = π?

**Solution:**

A function is said to be continuous when the graph of the function is a single unbroken curve.

The given function is

f(x) = x^{2 }− sin x + 5

It is evident that f is defined at x = π.

At x = π, f(x) = f(π)

= π^{2} − sinπ + 5

= π^{2} − 0 + 5 = π^{2} + 5

Consider lim_{x→π} f(x) = lim_{x→π} (x^{2} − sin x + 5)

Put x = π + h,

it is evident that if x→π, then h→0

⇒ lim_{x→π} f(x) = lim_{x→π} (x^{2 }− sin x) + 5

= lim _{h→0} [(π + h)^{2} − sin(π + h) + 5]

= lim _{h→0} (π+h)^{2} − lim _{h→0} sin (π+h) + lim _{h→0} 5

= (π + 0)^{2} − lim _{h→0} [sin π cos h + cos π sin h] + 5

= π^{2} − lim _{h→0} sin π cos h − lim _{h→0} cos π sin h + 5

= π^{2} − sin π cos 0 − cos π sin 0 + 5

= π^{2} − 0(1) − (−1)0 + 5

= π^{2} + 5 = f(π)

Therefore, the given function f is continuous at x = π

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.1 Question 20

## Is the function defined by f(x) = x^{2 }− sin x + 5 continuous at x = π?

**Summary:**

Hence we have concluded that the function defined by f(x) = x^{2 }− sin x + 5 continuous at x = π

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