# Define continuity of a function at a point x = a. Show that the function $$f(x) = \sqrt{1-x^2}$$ is   continuous on the interval[1, -1].

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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

1. f(a) is defined (i.e. a is in the domain of f)
2. $$\lim_{x \to a} f(x)$$ exists
3. $$\lim_{x \to a} f(x) = f(a)$$

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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