Limits and continuity test questions

In summary, the limits and continuity test is used to determine if a function is continuous at a specific point or over a given interval. To find the limit of a function, you can use various techniques such as substitution, factoring, or algebraic manipulation. A removable discontinuity occurs when a function has a hole or missing point at a specific value, but the limit of the function still exists. A non-removable discontinuity is when the limit of the function does not exist at a certain point due to an asymptote or jump in the graph. A function is continuous at a specific point if the limit of the function at that point exists and is equal to the value of the function at that point. It is possible for a function
  • #1
TheLostOne
1
0

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.
 
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  • #2
You can show 2a and 2b directly with the epsilon-delta-definition of continuity. So calculate what is f(x)-f(y) and show that this can be made arbitrarily small by putting x close to y.

For 3, your strategy is sound. You can see in 3b that the limit is +infinity from one side and -infinity from another.

4 is correct.

For 5, you can probably see that the functions may be discontinuous in a single point. It may help if you just plot the functions and see when they cross at the correct point. So for example in a), you'd plot f(x) = x+1 and g(x) = x^2 and see where they intersect.
 

1. What is the purpose of the limits and continuity test?

The limits and continuity test is used to determine if a function is continuous at a specific point or over a given interval. It helps to identify any potential discontinuities or breaks in the graph of a function.

2. How do you find the limit of a function?

To find the limit of a function, you can use various techniques such as substitution, factoring, or algebraic manipulation. If these methods do not work, you can use the limit laws or L'Hopital's rule to evaluate the limit.

3. What is the difference between a removable and non-removable discontinuity?

A removable discontinuity occurs when a function has a hole or missing point at a specific value, but the limit of the function still exists. A non-removable discontinuity, on the other hand, is when the limit of the function does not exist at a certain point due to an asymptote or jump in the graph.

4. How do you determine if a function is continuous at a specific point?

A function is continuous at a specific point if the limit of the function at that point exists and is equal to the value of the function at that point. This means that the graph of the function has no breaks or holes at that point.

5. Can a function be continuous but not differentiable?

Yes, a function can be continuous but not differentiable at a certain point. This occurs when there is a sharp turn or corner in the graph of the function at that point, making it impossible to find a derivative at that point.

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