Is the function f defined by
\[f(x) = \begin{cases} x, & \quad \text{if } n{\leq1}\\ 5, & \quad \text{if } n{ >1} \end{cases}\]continuous at x=0? At x=1? At x=2?
Is the function f defined by
\[f(x) = \begin{cases} x, & \quad \text{if } n{\leq1}\\ 5, & \quad \text{if } n{ >1} \end{cases}\]continuous at x=0? At x=1? At x=2?
Solution and Explanation
The given function f is \(f(x) = \begin{cases} x, & \quad \text{if } n{\leq1}\\ 5 & \quad \text{if } n{ >1} \end{cases}\)
At x = 0,
It is evident that f is defined at 0 and its value at 0 is 0.
Then, \(\lim\limits_{x \to 0}\) f(x) = \(\lim\limits_{x \to 0}\) x = 0
∴\(\lim\limits_{x \to 0}\) f(x) = f(0)
Therefore, f is continuous at x = 0
At x = 1,
f is defined at 1 and its value at 1 is 1.
The left hand limit of f at x=1 is,
∴\(\lim\limits_{x \to 1^-}\) f(x) = \(\lim\limits_{x \to 1^-}\)x=1
The right hand limit of f at x = 1 is,
∴\(\lim\limits_{x \to 1^+}\) f(x) = \(\lim\limits_{x \to 1^+}\)+(5) = 5
∴\(\lim\limits_{x \to 1^-}\) f(x) ≠ \(\lim\limits_{x \to 1^+}\)f(x)
Therefore, f is not continuous at x = 1
At x = 2,
f is defined at 2 and its value at 2 is 5
Then, \(\lim\limits_{x \to 2}\) f(x) = \(\lim\limits_{x \to 2}\) (5) = 5
∴\(\lim\limits_{x \to 2}\) f(x) = f(2)
Therefore, f is continuous at x=2
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.