1. Oct 21, 2008

### asdfsystema

I uploaded it as an image for easier read. Thank you for all the help !!!

2. Oct 22, 2008

### Tedjn

Your answers to 1. are not correct; you can confirm by graphing the function on a graphing calculator. It is not true that the fraction has a limit of infinity or negative infinity just because it is being divided by an expression containing x. What matters is how fast the numerator and denominator are increasing relative to each other. In this case, we see that the denominator has a factor of x. The numerator has a factor of x2 under a square root, so roughly a factor of x also. From this quick analysis, which you develop from experience, we can guess that the limit is going to be an actual number (roughly, the x's "cancel" to leave some real number behind).

How would we find the limit in this case? The square root is a problem, because we can't break it up. One way to take care of this is to simply put the bottom under a square root as well, so that you can combine the numerator and denominator under a huge square root. Try this, and play with it.

For your number 2, recheck your answer. What is the sign of the denominator? What is the sign of the numerator?

For your number 3, only your third one is correct. In fact, a limit of positive or negative infinity would imply that those functions do not have a horizontal asymptote. Why? Explain your through process to us so that we can see where you are going wrong. For the fourth and fifth ones, you can do something similar to 1.

4. f(s) is piecewise, and each piece is continuous (very simply, because s2-c and cs+1 are polynomials). The only part where the function might not be continuous is at the point where the two branches meet, at s = 8. What must be true about the two branches of the function when s = 8 for the function to be continuous? Why?

5. Yes, you are right that you must redefine f(-3) = 2 for the function to be continuous. However, why would the function be continuous if that is true? You can try graphing it to see what happens if f(-3) is not equal to 2. Finally, remember that you must also show that there is a removable discontinuity in the first place at x = -3. Since f(-3) is defined, you must then show that the function is not continuous at x = -3. This should be simple if you have a clear understanding of why f(-3) must equal 2.

Last question. Unfortunately, the answer is not 6. How did you reach that conclusion? This question is actually a little tricky. You might want to start by first writing tan6x in terms of sine and cosine. Then, you can use the fact that sin(y)/y = 1 as y -> 0.

3. Oct 22, 2008

### asdfsystema

1)
a) lim -> inf answer: 10/4?
b) lim-> neg inf answer: 10/4 ?
still unsure…
I think a and b will have same answers

2) ah I see what I did wrong
a) negative infinity
b) negative infinity
c) negative infinity

3)
a) –4 ? I just did –8/2 since 6 doesn’t matter if its going to infinity which gives me –4
b) 0 ? very large denominator, very small numerator
c) –1/3
d) – 1/3 , but is there a quicker method ? I had to use calculator and I want to try without using calculator
e) 1/3 same as d, but I used calculator, is there a fast method

4)the functions cannot equal to 8 where the piecewise is ? what are the next steps to this ?

5) I’m not supposed to use a calculator but I did it anyways for the sake of solving this problem. Wait, is the answer =2 ? when I graphed it, its basically a frowning and smiling parabola with y=1. if f(-3)=2 , then it’s removable discontinuity? So 2 should be the answer?

6) sorry I meant 3. I’m pretty sure that is correct ?

Thank you for responding really quick ! I appreciate it. Still a newbie at this…

Sorry if it is very hard to read, I tried to use different colors so its easier. Thanks again

4. Oct 22, 2008

### Tedjn

1. Still not quite, but you are getting closer. Don't forget about the square root. The two limits will not be the same, which you will see if you graph it, but the graph does possess some symmetry.

2. Very good.

3. a, b, c) All good.

3. d, e) This question is identical to #1. My suggestion was to put everything underneath the square root sign first, and then take a square root at the end. Do you understand what I mean?

4. The intuitive idea is that the two branches must meet at the point which switches from one branch to the other. For x < 8, we follow the first branch. At x = 8, we suddenly jump to the second branch. At this point x = 8, if the two branches don't meet, then the function isn't continuous. What if they do meet?

5. The answer is -2, but the question also asks you to show that the discontinuity is removable. That is, after defining f(-3) = 2, you need to show that the function is now continuous. It is similar to #4.

6. Yes, I believe 3 is correct.

5. Oct 23, 2008

### asdfsystema

eh i'm a retard ...

1. a) sqrt(10)/4
b) negative sqrt(10) /4

3. d) i keep getting -1/3 because for example I plug in 100 , sqrt (11200)/ -300 = 105/-300 = -1/3
e) . plugging in -100-> sqrt (8000)/ 300 -> 89/300 hmm I get 0 if I keep going bigger

4. if they do meet, then that means they will be continous at that point then and we can use the equation since it can now be greater or equal to / lesser than or equal to ?

5. with the answer being -2, do I have to do any other steps to prove it ? or -2 suffices?

Thanks

6. Oct 23, 2008