Limits

To find the equation of the normal to the ellipse at the point where the line is tangent, we proceed as follows:

1. Substituting the Line into the Ellipse:
The line is , so solve for :


Substitute into the ellipse equation :


Expand:



So:

2. Tangency Condition:
Since the line is tangent to the ellipse, there is exactly one point of intersection, so the quadratic in has one solution. The discriminant must be zero:

For , the discriminant is:


Set :

3. Finding the Point of Tangency:
With , the quadratic becomes:


Divide by 3:


Rewrite:



Find :


So, the point is .

4. Slope of the Tangent:
Differentiate the ellipse implicitly:



At :

5. Slope of the Normal:
The slope of the normal is the negative reciprocal of the tangent’s slope:

6. Equation of the Normal:
Using point-slope form at :


Multiply through by :





Thus:

Final Answer:
The equation of the normal to the ellipse at point is .

Step 1: Check continuity at
For to be continuous at , we must have: Since , we evaluate: Using the approximations: Substituting, Taking the limit as , Thus, for continuity,

Step 2: Compute the derivative at
The derivative at is given by: Using the same approximations: Since , we apply L'Hôpital's Rule to: Differentiating numerator and denominator:

Step 3: Conclusion
Thus, the correct answer is:

Step 1: Analyze the function behavior
The given function is: Using the definition of the modulus function, we consider two cases: 1. Case 1: When Substituting in : Hence, for , is a constant function with value 1. 2. Case 2: When Substituting in : Hence, for , is a constant function with value -1.

Step 2: Check continuity at
The function is undefined at because the denominator becomes zero. This means is discontinuous at .

Step 3: Conclusion
- The function is not continuous at . - The function is constant (equal to 1) for . - The function is also constant (equal to -1) for . - The function does not strictly increase or decrease for . Thus, the correct answer is:

Step 1: Compute
Substituting in :

Step 2: Compute using Taylor Expansion
Expanding using first-order approximations: Approximating terms using for small : For cosine, Substituting these approximations:

Step 3: Compute the difference

Step 4: Compute the limit
Substituting: Factor from the numerator and denominator: Cancel : Substituting :

Step 5: Conclusion
Thus, the correct answer is:

Step 1: Understanding the Given Expression
We are given the expression: where denotes the fractional part of , i.e., . As , we know that since the fractional part of becomes very small as approaches 0 from the positive side. We are asked to find the value of .

Step 2: Approximation of Functions
As , we have . Therefore, we can use small angle approximations for the inverse trigonometric functions. For small values of , we know: and Thus, for being small, we have: and

Step 3: Substituting the Approximations into the Expression
Now, substitute these approximations into the given expression: This simplifies to:

Step 4: Simplifying the Expression
For small , we can neglect in comparison to . Therefore, the denominator simplifies to . The expression now becomes: Thus, we have:

Step 5: Final Calculation
We are asked to find . Since , we compute: Therefore, the value of is .

Step 6: Final Answer
The correct answer is: