Limits

The given function is: Step 1: First, find : As , both and behave such that the first term approaches infinity and the second approaches 0. Thus, .
Step 2: Now, we calculate , giving the value of to be .
Thus, the set of all possible finite values of and is .

Evaluating the limit To check continuity at : Multiplying numerator and denominator by the conjugate: Since , we simplify: Substituting :

Step 1: Approximate the Square Root Expansions
Using the first-order binomial approximation: Expanding : Thus, expanding the nested square root term: Applying binomial approximation again:
Step 2: Approximate the Second Square Root Term
Expanding : Using the binomial expansion: Thus,
Step 3: Compute the Limit
Now, the numerator simplifies to: Rearranging: For small , the dominant term in the numerator is: Thus, the limit evaluates to:
Final Answer:

Step 1: Apply First-Order Approximations
Using the first-order Taylor series expansion near :
Step 2: Simplify the Numerator and Denominator
The numerator: The denominator: Expanding the square:
Step 3: Compute the Limit
Dividing the numerator by the denominator:
Final Answer:

Step 1: Apply the Continuity Condition at
For to be continuous at , we must have: Thus, we evaluate the left-hand limit and right-hand limit separately.
Step 2: Evaluate Right-Hand Limit (RHL)
Using the approximations for small , we get: So, .
Step 3: Evaluate Left-Hand Limit (LHL)
Using approximations and for small , we get: So, .
Step 4: Compute

Final Answer: .

Step 1: Use standard limits for trigonometric functions as .
We will use the following well-known limits:
1.
2.
Step 2: Rewrite the numerator using standard limit forms.
The numerator is .
For the first term, .
As , this term behaves like . For the second term, .
As , this term behaves like . So, the numerator approximately becomes as . Step 3: Rewrite the denominator using standard limit forms. The denominator is .
We know . So, as .
Therefore, as .
Step 4: Substitute the simplified expressions back into the limit. Cancel out : Now simplify the fraction. Both 1305 and 81 are divisible by 9.

Step 1: Analyze the given limit expression and identify indeterminate forms.
The given limit is . Let's evaluate the numerator as :
The term .
As , and . This part is an indeterminate form.
The term . As , .
So, the entire numerator approaches . Now, let's evaluate the denominator as :
The denominator is .
As , .
So the denominator approaches .
Since both the numerator and the denominator approach 0, the limit is of the indeterminate form.
Step 2: Simplify the numerator using standard limit forms.
Let the numerator be . Rewrite the terms to utilize standard limits as : Multiply and divide by appropriate powers of to get standard limit forms: Recall the following standard limits as :



Using these, the numerator effectively behaves like as .
Step 3: Simplify the denominator by rationalizing.
Let the denominator be .
Since approaches 0 as and contains square roots, rationalize it by multiplying the numerator and denominator by its conjugate, : As , the term . So, the denominator effectively behaves like as .
Step 4: Substitute the simplified forms back into the limit and evaluate.
Now, substitute the simplified forms of the numerator and denominator back into the limit expression: Rearrange the terms: We know that . Therefore, . Substitute this value into the limit expression:

For the function to be continuous at , the following condition must be satisfied: We are given .
Step 1: Evaluate the left-hand limit ( ).
For , . This limit is of the indeterminate form . We can evaluate it using the formula: .
Here, and .
So, from the continuity condition, we have .
Step 2: Evaluate the right-hand limit ( ).
For , .
As , . In the expression, grows faster than .
To evaluate the limit, divide the numerator and the denominator by : As , , which means . Therefore, .
Substituting this into the limit expression: So, the right-hand limit is .
Step 3: Equate the limits and to find and .
From the continuity condition: From this, we have two equations:
1.
2.
Step 4: Calculate the value of . The final answer is .