I've got quite a few problems. Please answer any of them in whole or in part. Thank you in advance. (For the record, you're not doing my homework; you're helping me understand it. These problems were not assigned.)

(1). Determine the discontinuities and whether or not they're removable (if they are, remove them) where f(x) = sin x if x is less than or equal to pi over 4, and f(x) = cos x if x > pi / 4.

(2). Is this continuous at all points? I'd say it's not continuous at x = 3, 2 because the denominator is factorable. Are these removable discontinuities? Am I even close to right?The equation: f(x) = (x^3 - 2x +7) / (x^2 - 5x + 6)

(3). f(x) = 2x^2 - 5x - 3 if x does not equal 3; 6 if x = 3. Find points of discontinuity and remove them if they're removable.

(4). Name the points at which the function is discontinuous and tell whether or not it's removable. f(x) = (x-1) / (x^2 + 1)

(5). Find the limit as x approaches negative 1 of sinx / x + 1

(6). Find the limit as x approaches pi for the equation sinx / (2 + cos x).

(7). Find the limit as x approaches zero from the left and from the right of the equation ((1/x) - (1/abs(x))). That is, one over x minus one over the absolute value of x.

Thanks!

That's a lot more than "a few".

(5) Lim (x-> -1) sinx /(x + 1) = infinity, because the mumerator remains finite while the denominator goes to zero

(6)limit x ->pi sinx/(2+cos x).
= sin pi/(2 + cos pi)
= 0/1 = 0

That's exactly why I said "quite" before I said "a few." Thanks for your help.

(1). To determine the discontinuities of the function f(x) = sin x for x ≤ π/4 and f(x) = cos x for x > π/4, we need to consider where the function changes its definition or behavior.

The given function has a piecewise definition based on the value of x. At x = π/4, there is a point of discontinuity because the function changes from sine to cosine. However, this point is not a removable discontinuity because neither sine nor cosine can be extended continuously.

Therefore, the only discontinuity in f(x) is at x = π/4, and it is not removable.

(2). To determine if the function f(x) = (x^3 - 2x + 7) / (x^2 - 5x + 6) is continuous at all points, we need to check if there are any points where the function is undefined or behaves differently.

The denominator of the function can be factored as (x - 3)(x - 2). This means that the function is undefined at x = 3 and x = 2 because these values would make the denominator equal to zero.

To determine if these points are removable discontinuities, we need to check if the numerator also evaluates to zero at those points. If the numerator is zero at those points, it would imply a common factor, which can be canceled out to remove the discontinuity.

To find out if x = 3 and x = 2 are removable discontinuities, substitute these values into the numerator and see if they result in zero:

f(3) = (3^3 - 2(3) + 7) = 34
f(2) = (2^3 - 2(2) + 7) = 11

Since the numerator does not equal zero at x = 3 and x = 2, these points are not removable discontinuities. Therefore, the function is discontinuous at x = 3 and x = 2.

(3). The function f(x) is given as f(x) = 2x^2 - 5x - 3 for x ≠ 3 and f(x) = 6 for x = 3.

To find the points of discontinuity and remove them if possible, we need to check if f(x) is continuous at x = 3.

For x ≠ 3, the function f(x) does not have any apparent discontinuities.

At x = 3, f(x) is given as 6. This means that the value of the function is defined and continuous at x = 3.

Since there are no other apparent points of discontinuity, we can conclude that the function f(x) is continuous everywhere except at x = 3, where it has a removable discontinuity.

(4). The function f(x) is given as f(x) = (x - 1) / (x^2 + 1).

To find the points of discontinuity, we need to look for values of x that cause the denominator to equal zero, as those would make the function undefined.

Setting the denominator equal to zero:

x^2 + 1 = 0

This equation has no real solutions, which means there are no values of x that make the denominator zero. Therefore, the function f(x) is continuous for all x and does not have any points of discontinuity.

(5). To find the limit as x approaches -1 of sin x / (x + 1), we can use direct substitution.

Plugging in -1 for x:

lim(x→-1) sin x / (x + 1) = sin(-1) / (-1 + 1) = sin(-1) / 0

Since the denominator is zero, the limit does not exist.

(6). To find the limit as x approaches π for the function sin x / (2 + cos x), substitute π into the expression:

lim(x→π) sin x / (2 + cos x) = sin(π) / (2 + cos π) = 0 / (2 + (-1)) = 0 / 1 = 0

Therefore, the limit as x approaches π for the given function is 0.

(7). To find the limit as x approaches 0 from the left and from the right of the expression (1/x) - (1/|x|):

From the left side (x → 0-):

lim(x→0-) (1/x) - (1/|x|) = (1/0) - (1/|0|) = ∞ - ∞

This is an indeterminate form and requires further analysis. The limit from the left is undefined.

From the right side (x → 0+):

lim(x→0+) (1/x) - (1/|x|) = (1/0) - (1/|0|) = ∞ - ∞

Again, this is an indeterminate form and requires further analysis. The limit from the right is undefined.

Therefore, the limit as x approaches 0 from both sides of the expression (1/x) - (1/|x|) is undefined.