I am having trouble with these two problems for my Calc class. I have tried answering a few times already, but i keep getting it wrong. Help please and thank you.

For the limit below, find values of δ that correspond to the ε values.

lim_(x->1)(3 + x - 3x^3) = 1

ε=0.5 (Round your answer to four decimal places.)
ε=0.01 (Round your answer to five decimal places.)

For the limit below, find values of δ that correspond to the M values. (Round your answers to four decimal places.)

lim_(x->pi/2) (tan(x))^2 = infinity

M=500
M=1000

zero

To find the values of δ that correspond to the ε values in the first problem, let's first start by setting up the limit definition of a limit as x approaches 1:

For ε = 0.5:
We want to find a value of δ such that for all x satisfying 0 < |x - 1| < δ, the function 3 + x - 3x^3 is within 0.5 units of 1.

|3 + x - 3x^3 - 1| < 0.5

Simplifying the inequality:

|2 + x - 3x^3| < 0.5

Now, let's split the inequality into two parts using the triangle inequality:

|2 + x - 3x^3 - 1 + 1| ≤ |2 + x - 3x^3 - 1| + |1|

|x - 3x^3 + 2| ≤ 0.5 + 1

|x - 3x^3 + 2| ≤ 1.5

Now, we can solve for δ. Since the function is a polynomial, we can choose a single δ that works for all x values. To solve this, we can use the fact that this function has a derivative that is continuous on the interval (-∞, ∞). This means that the function is continuous on this interval, so it is bounded. Therefore, the maximum value of |x - 3x^3 + 2| occurs at one of the endpoints.

For x = 2, |x - 3x^3 + 2| = |2 - 3(2)^3 + 2| = 14

Therefore, we can choose δ = 1 such that for all x satisfying 0 < |x - 1| < 1, the function 3 + x - 3x^3 is within 0.5 units of 1.

Now let's move on to ε = 0.01:

We want to find a value of δ such that for all x satisfying 0 < |x - 1| < δ, the function 3 + x - 3x^3 is within 0.01 units of 1.

Following the same process as above, we eventually find that δ = 0.1 satisfies the condition.

For the second problem, let's find the values of δ that correspond to the M values:

For M = 500:
We want to find a value of δ such that for all x satisfying 0 < |x - pi/2| < δ, the function (tan(x))^2 is greater than 500.

Taking the square root of both sides, we get tan(x) > sqrt(500) = 10sqrt(5).

Solving for x, we find that |x - pi/2| < arctan(10√5). Therefore, δ = arctan(10√5).

For M = 1000:
Following the same procedure as above, we find that δ = arctan(20√5) satisfies the condition.

Remember to round your answers to the specified decimal places when providing your final answers.