Prove that the function f(x) = 5x-3 is continuous at x = 0, at x = -3 and at x = 5.
Prove that the function f(x) = 5x-3 is continuous at x = 0, at x = -3 and at x = 5.
Solution and Explanation
The given function is f(x) = 5x-3
At x = 0, f(0) = 5x0-3 = 3
\(\lim\limits_{x \to 0}\) f(x) = \(\lim\limits_{x \to 0}\)(5x-3) = 5x0-3 = -3
∴\(\lim\limits_{x \to 0}\) f(x) = f(0)
Therefore,f is continuous at x=0
At x=-3, f(-3) = 5x(-3)-3 = -18
∴\(\lim\limits_{x \to 0}\) = \(\lim\limits_{x \to 3}\) (5x-3) = 5x(-3)-3 = -18
∴f(x) = f(-3)
Therefore, f is continuous at x=−3
At x=5, f(5) = 5x(5)-3 = 22
∴\(\lim\limits_{x \to 5}\)=\(\lim\limits_{x \to 5}\) (5x-3) = 5x(5)-3= 22
∴f(x) = f(5)
Therefore, f is continuous at x = 5
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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.