- #1
TheLostOne
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Homework Statement
2. Show that the function is continuous on the given interval.
(a)f(x)= (2x+3)/(x-2) range:(2, infinity)
(b)f(x) = 1- sqrt(1-x^2) range:[-1,1]
3. Prove that the following limits do not exist.
(a) lim x tends to 0 ( absolute|x|/x)
(b) lim x tends to 3 (2x/(x-3))
4.Find the limit of the function
f(x) = x^2 if x=/= 0
2 if x=0
at the point (a) x=-1; (b) x=0; (c) x=sqrt(2)
5. Find all values of a such that f is continuous on (-infinity, infinity):
(a)
f(x)= x+1 if x<=a
x^2 if x>a
(b) f(x) = (x^2 + a)/(x-1) if x=/=1
2 if x =1
The Attempt at a Solution
Can I answer question 2a and 2b by writing out statements to prove that they are continuous? If that is not allowed, how do I prove that they are continuous?
I know 2a is continuous on (2, infinity) because 2x + 3 and x - 2 are polynomials and the only point of discontinuity occurs at x=2 as that will cause the denominator to be zero. I also know that for 2b, lim x tends to a f(x) =f(a) when -1<=a<=1 so f(x) is continuous on [-1, 1]. But how do I SHOW it on the paper?
For question 3a, Can I show that left hand side limit = -1 while right hand side limit = 1, and since left hand side limit is not = to right hand side limit, the limit doesn't exist. For question 3b, do I do the left hand and right hand limit test again or do I just substitute in x=3?
For question 4, I believe I should just substitute in the x-values when x=-1 and x= sqrt(2). To find the limit for x=0, I could find the left and right hand limit to conclude that lim x tends to 0 is equal to 0. I think that's how it should be done, but it seems too easy to be true.
As for question 5, I don't have any ideas on how I should start or what I should do, so I hope that someone could guide me on that.
Many thanks.
P.S. If my attempt is not informative enough or unsatisfactory, I guess I could try writing it more mathematically.