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
Top Questions on Continuity and differentiability
- If the function
\[ f(x) = \begin{cases} \sin(2x), & \text{for } x > 0, \\ a + bx, & \text{for } x \leq 0, \end{cases} \]
where \( a \) and \( b \) are constants, is differentiable at \( x = 0 \), then \( a + b \) is equal to:- GATE BT - 2025
- Mathematics
- Continuity and differentiability
- If $$ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6, \text{ then } f(1) \text{ is equal to:} $$
- JEE Main - 2025
- Mathematics
- Continuity and differentiability
- Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice-differentiable function such that \( f(2) = 1 \). If \( F(x) = x f(x) \) for all \( x \in \mathbb{R} \), and the integrals \( \int_0^2 x F'(x) \, dx = 6 \) and \( \int_0^2 x^2 F''(x) \, dx = 40 \), then \( F'(2) + \int_0^2 F(x) \, dx \) is equal to:
- JEE Main - 2025
- Mathematics
- Continuity and differentiability
- For $a, b > 0$, let $ f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x < 0, \\ \frac{x}{3}, & x = 0, \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b\sqrt{a x \sqrt{x}}}, & x > 0 \end{cases} $ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to:
- JEE Main - 2024
- Mathematics
- Continuity and differentiability
The function \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} 2+x, & \text{if } x \geq 0 \\ 2-x, & \text{if } x \leq 0 \end{cases} \] Then function \( f(x) \) at \( x=0 \) is:
- IIITH UGEE - 2024
- Mathematics
- Continuity and differentiability
Notes on Continuity and differentiability
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.