Limits

We need to evaluate the integral: The expression , and we can use substitution. Let , hence . Thus, the integral simplifies as: Thus, the correct answer is (C).

We are given the differential equation: Step 1: Rearranging the equation
Divide both sides by to isolate : Step 2: Separate the variables
Actually, let's directly rearrange: Step 3: Integrate both sides
Recall that: So: Step 4: Exponentiate both sides
Now integrate: Thus, the required solution is:

Given: , and
Step 1: Use trigonometric identities for sum
Step 2: Eliminate
Let us square and add both expressions: Step 3: Use expression for
Now divide both the original expressions: Use:

We are given the integral: First, factor out from the expression under the square root: Thus, the integral becomes: Now, let , then the integral simplifies as: Thus, the correct answer is (C).

Step 1: Given equation is
Step 2: Replace with in equation (1)
Step 3: Solve equations (1) and (2) simultaneously
Equation (1):
Equation (2): Multiply equation (1) by 2:
Now subtract (2) from (3):
Put this value back into equation (1):

Step 1: Simplify the expression

The given limit is: We will evaluate this limit as . - First, observe that grows very quickly as . This suggests that dominates both the numerator and the denominator for very small values of . - The term can be approximated as follows: since is a constant and . Similarly, can be approximated as: because is also a constant.

Step 2: Analyze the limit

For very small , the expression simplifies to: As , grows very large. So, the expression becomes approximately: However, we are given that the limit is equal to 2. This means that the numerator and denominator must approach each other in such a way that their ratio is 2. To ensure this, we must adjust the coefficients such that the ratio of the terms balances correctly.

Step 3: Find the values of and

In order to match the desired limit of 2, we need to determine the values of and that balance the terms. Given the structure of the problem, a logical assumption would be to set and , as these choices will yield the correct balance between the terms. Therefore, we have:

Final Answer:

The values of and are and .

Step 1: Express the number in a simpler form

We are asked to find the approximate value of , which is the cube root of 0.007. We can rewrite this as: This can be broken into: Since , we now need to approximate .

Step 2: Approximate the cube root of 7

The cube root of 7 is approximately . Therefore:

Step 3: Multiply the results

Now, we multiply the approximate values:

Final Answer:

The approximate value of is .

We start with the given limit: We can rewrite this as: The first limit is well known: For the second term, since as , we have: Therefore, the limit is: Thus, , so . The value of can be any real number because it does not affect the limit.

Let
Multiply numerator and denominator by the conjugate of denominator:

Now find the argument:
But the complex number lies in the third quadrant Argument =

The expression given is , and we need to find its value.

Step 1: Analyze the powers of complex numbers

Let and . Notice that these are conjugates of each other, and the sum of their powers will yield a real number since the imaginary parts cancel each other out when added. Let’s express both and in polar form.

The modulus of and is:

The argument of is: The argument of is: So, we can write the polar forms of and as: and Using De Moivre's Theorem, we raise these to the power of 100: and Since and , adding these two expressions cancels out the imaginary parts, leaving only the real parts: Now, simplifying the angle: So, Thus,

Step 2: Add

Now, the full expression is: Finally, the sum is:

Final Answer:

The value of is .

We are given the second-order differential equation: Step 1: Integrate once with respect to
Step 2: Integrate again
Step 3: Analyze the result
The solution is the general equation of a straight line.
Therefore, the differential equation represents all straight lines in a plane.

Recall: , so powers of repeat every 4 terms.

simplify:

Final answer:

Step 1: Simplify the first term
Step 2: Simplify the second term
Step 3: Add the two simplified expressions
Step 4: Use identity
Step 5: Final simplification

We are given the expression . Let's break it down: 1. because the logarithm of a number to the same base is always 1. 2. Therefore, . Now, for , we apply the change of base formula: Since , this simplifies to: Finally subtracting the values: Thus, the correct answer is 0.

To check the continuity of the function at , we need to ensure that: From the problem, we know that: - - - We will now compute using the binomial expansion for : Thus: So: For continuity at , we need: Solving for : Thus, the correct answer is .

We are asked to find the limit: Step 1: Use the logarithmic approximation for small . For any constant , we know the approximation for when is near 0: Using this approximation for and , we get: Step 2: Substitute into the limit. Substitute these approximations into the original expression: Simplifying: The 's cancel out: This simplifies to: Thus, the value of the limit is . Therefore, the correct answer is option (C)

We are given that and are distinct roots of the quadratic equation , which implies: We need to evaluate the limit: Since at , we have: Now, we expand around using a Taylor expansion. Let . Near , we use the first two terms of the expansion: where (since is a root), and Thus, at , Now, we have: Using this expansion in , we get: Substituting , we get: Thus: Now, substituting this into the original limit expression, we have: Using the fact that and are distinct roots, we conclude that the correct value for the limit is: Thus, the correct answer is .