Question

Is \[ f(x)=x^2 \] continuous at \(x=0\)?

• A. Yes
• B. No
• C. Only from left
• D. Only from right

Hint:

All polynomial functions are continuous for every real value of \(x\).

Answer 1

The Correct Option is A

Concept: A function \(f(x)\) is continuous at \(x=a\) if \[ \lim_{x\to a}f(x)=f(a) \] That is, \[ \text{LHL}=\text{RHL}=f(a) \] must hold simultaneously.

Step 1: Find the value of the function at \(x=0\) Given \[ f(x)=x^2 \] Substituting \(x=0\), \[ f(0)=0^2=0 \] \[ f(0)=0 \]

Step 2: Find the left-hand limit \[ \lim_{x\to0^-}x^2 \] As \(x\) approaches \(0\) from the left side, \[ x^2\to0 \] Therefore, \[ \lim_{x\to0^-}f(x)=0 \]

Step 3: Find the right-hand limit \[ \lim_{x\to0^+}x^2 \] As \(x\) approaches \(0\) from the right side, \[ x^2\to0 \] Hence, \[ \lim_{x\to0^+}f(x)=0 \]

Step 4: Compare the limits and function value We obtain \[ \text{LHL}=0 \] \[ \text{RHL}=0 \] and \[ f(0)=0 \] Thus, \[ \lim_{x\to0}f(x)=f(0) \] Therefore, the function is continuous at \(x=0\). Final Answer: \[ \boxed{\text{Yes}} \]