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].

Answers

This answer is not selected as best answer. This answer may not be sufficient for exam.

Your limit has been exceed. We have implemented this system because, We got difficulty on managing our servers. Please donate some amount to remove this limit.

Quota: 0 / 30

Donate

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]

If you found any type of error on the answer then please mention on the comment or submit your new answer.
Leave your Answer:

Click here to submit your answer.

Discussion
0 Comments
  Loading . . .