Limit And Continuity

We want to evaluate:

This can be rewritten as:

The second term is simply .

Let's focus on the first term:

This limit is of the indeterminate form . We can use L'Hopital's rule, but it's easier to use the fact that .

Let . Then as , . Also, note that as .

Since and , then the entire limit will be .

Therefore:

Step 1: Analyze the piecewise function at critical points and :

Step 2: Check continuity at : Since all three values match, is continuous at .

Step 3: Check continuity at : Since , is not continuous at .

Step 4: Verify continuity elsewhere: - For : is continuous - For : is continuous - For : is continuous

Conclusion: The function is discontinuous only at , making option (D) correct:

Let's simplify the expression inside the parentheses:

Now, as x approaches 0:

Now, let's factor out from the expression:

Now, as x approaches 0:

So, the correct option is (C): 1

Step 1: Substitute Substituting directly: Since both numerator and denominator become zero, apply L'Hôpital's Rule.
Step 2: Differentiate numerator and denominator Differentiate the numerator: Differentiate the denominator:
Step 3: Compute the limit Substituting : Thus, the correct answer is:

We are given the following limit: We need to evaluate this limit as .
Step 1: Use the approximation for small , which is when is close to 0. Therefore, for small , we have: Thus, .
Step 2: Now, substitute into the sine function: For small , we can use the approximation when is small. Therefore: Step 3: Substituting this approximation into the original limit expression, we get: Thus, the correct answer is option (E).