Continuity

Step 1: Continuity condition at . For to be continuous at , the left-hand limit ( ) must equal the right-hand limit ( ) and the value of . Equating and : Step 2: Solving the equation. Rearranging: Conclusion: Thus, the required value of is , which corresponds to option .

For the function to be continuous at , the limit of as approaches 0 must equal the value of the function at . Therefore, we need to evaluate the limit: We know that as , so . Therefore: Thus, the limit becomes: For the function to be continuous at , this limit must equal the value of . Therefore: So, the correct value of is .

The function is given as: To check continuity at , we need to check if . We know that oscillates between -1 and 1 as approaches 0. Hence: Multiplying by , we get: As , both and approach 0. By the Squeeze Theorem, we conclude that: Thus, is continuous at . The assertion is correct, and the reason is also correct. Therefore, both the assertion and reason are true.

To check continuity at , we need to verify that the left-hand and right-hand limits at are equal to the function value at . - The left-hand limit at is . - The right-hand limit at is . Since the left-hand and right-hand limits are not equal, the function is not continuous at .

For the function to be continuous at and , the left-hand and right-hand limits must be equal. At , we have: Thus, . Solving for the second condition at : Solving these equations will yield and .

For the function to be continuous at , we must have We need to calculate the limit of as approaches . First, simplify the expression for when : Factor the numerator: Thus, For , the terms cancel out, and we are left with: Now, compute the limit as approaches : For continuity at , we need . Hence,