Question

Let\[f(x)=\begin{cases} 2x-1, x<1\\ 1, x=1 \\ x^2,x>1 \end{cases}\]then at x = 1

• A. f(x) is continuous from left only
• B. f(x) is continuous from right only
• C. f(x) is continuous
• D. f(x) has removable discontinuity

Answer 1

The Correct Option is C

The function \(f(x)\) is defined piecewise as follows:

\[ f(x)=\begin{cases} 2x-1, & x < 1 \\ 1, & x=1 \\ x^2, & x > 1 \end{cases} \]

To determine whether \(f(x)\) is continuous at \(x=1\), we must check the left-hand limit, the right-hand limit, and the value of the function at \(x=1\).

The left-hand limit as \(x\) approaches 1 is:

\[\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x-1) = 2(1) - 1 = 1\]

The right-hand limit as \(x\) approaches 1 is:

\[\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1^2 = 1\]

The value of the function at \(x=1\) is:

\[f(1) = 1\]

Since the left-hand limit, the right-hand limit, and the value of the function at \(x=1\) are all equal, we can conclude that:

\[\lim_{x \to 1} f(x) = f(1) = 1\]

Therefore, the function \(f(x)\) is continuous at \(x=1\).

The correct answer is: f(x) is continuous.