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**Definition:** A function f(x) is continuous at x = a if \(\lim_{x \to a} f(x) = f(a)\) [Functional Value = Limiting Value]

Precisely, a function f(x) is continuous at x = a if

- f(a) is defined (i.e. a is in the domain of
*f*) - \(\lim_{x \to a} f(x)\) exists
- \(\lim_{x \to a} f(x) = f(a)\)

**Problem Part:**

Solution:

Given

\(f(x) = \sqrt{1-x^2}\)

\(a \in (-1, 1) \enspace i.e -1 < a < 1 \enspace then\)

\(\lim_{x \to a} f(x) = \lim_{x \to a}(\sqrt{1 – x^2})\)

\(= \lim_{x \to a} \sqrt{1 – x^2})\)

\(= \sqrt{1 – a^2}\)

= f(a)

Which shows that f(x) is continuous at \(x = a \in (-1, 1)\).

For the end points i.e. x = -1

\(\lim_{x \to -1^{+}}f(x) = \sqrt{1 – (-1)^2} = 0 = f(1)\)

Which shows that f(x) is continuous at left end point x = -1

For the end points i.e. x = +1

\(\lim_{x \to +1^{-}}f(x) = \sqrt{1 – (1)^2} = 0 = f(1)\)

Which shows that f(x) is continuous from the left at x = 1

Hence f(x) is continuous at [-1, 1]

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