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