Limits And Exponential Functions

To evaluate the given limit, we first express it mathematically:

The expression involves exponential terms, and it is beneficial to factor out the highest powers of from both the numerator and the denominator. Let's perform the simplification step-by-step:

1. Notice that the dominant term inside the exponential functions for both numerator and denominator is or , respectively.
2. Rewriting each component with base simplification:
   • For the numerator:
   • For the denominator:
3. We start with approximating the dominant factors:
4. For the polynomial terms:
   • In the numerator: factor out from . Approximately, is the dominant term.
   • In the denominator: factor out from . Approximately, is the dominant term.

Substitute these back into the original limit:

Canceling out the common terms from both the numerator and the denominator, we simplify the expression to:

Therefore, the value of the limit is:

(option D).

We are given the limit:

To solve for , we first use the fact that:

So the limit simplifies to: Thus, . Using the approximation , we find that .

Therefore, the value of is approximately 18280.

Concept: Use approximation: Also, conditions for both roots of a quadratic to lie in can be obtained using interval root conditions.

Step 1: {Evaluate the limit.} Thus, For finite limit numerator must vanish at : Condition satisfied. Now differentiate numerator: At : Given limit :

Step 2: {Form the quadratic.}

Step 3: {Apply root interval condition.} For both roots in : Also discriminant . Solving these inequalities gives

Step 4: {Find required value.}

Step 1: Understanding the Concept:
We first evaluate the limit to find the expression for and then solve the trigonometric equation.
: Key Formula or Approach:
Standard limits: and .
Step 2: Detailed Explanation:
.
Let . As , .
.
.
Equation to solve: .
Let .
(One solution is ).
. .
As , . By IVT, there is a root in .
Since is an odd function, if is a root, is also a root.
Checking number of intersections of and :
At , slopes are 0 and 2. is initially above.
Since eventually exceeds (which is bounded by 2), they must intersect once more in .
Total solutions: , one positive, and one negative. Total = 3.
Step 3: Final Answer:
The number of solutions is 3.