Asked by Mark
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!
(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!
Answers
Answered by
drwls
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
(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
Answered by
Mark
That's exactly why I said "quite" before I said "a few." Thanks for your help.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.