Question

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!

Answer 1

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

Answer 2

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