Remaining questions from Assignment #3

Apologies for not keeping track of the time better today- I'll do a better job of making sure that all homework questions are answered. So, here goes...

p. 72/19b, c, d
The limit exists if the function approaches *the same* y-value from both the left and right. This is true for all values of c between 0 and 1 and between 1 and 2. It is not true at c = 0, c = 1, or c = 2. At c = 0, the right hand limit exists (it is 1), but the left hand limit does not (think about function values like –0.1, –0.01, –0.001; they are all undefined). At c = 1, the left hand limit is 0 while the right hand limit is 1. At c = 2, the left hand limit is 1 (it is irrelevant that f(2) is 2).
Thus, the answer to part b is the intervals (0, 1) and (1, 2).
The answer to part c is c = 2 (since at c = 1, both the left and right hand limits exist).
Likewise, the answer to part d is c = 0.

p. 72/63
A nice way to think through this problem is to use some example functions and an example value of c. For example, if c = 1, f(x) = 2x - 1 has a limit of 1 and g(x) = x – 3 has a limit of –2. Now, you can write limit expressions for a, b, c, and d and see that all four are friendly functions at c = 1, and so all can be evaluated directly. For a more mathematical approach, take a look at Theorem 1 on p. 68. Voila!

p. 72/67
You know how to find this limit numerically (with a table) and graphically (like you did in #9), but to find it algebraically, you need a trick. How do you make it a friendly function if it doesn't factor? Well, fortunately, they give you the trick of multiplying by a clever FuFOO (Funny Form Of One). Here goes:







Now, the function in the limit has not changed (we've only multiplied by one, after all). What *has* changed, is that it is now obvious that an x can be cancelled from top and bottom, as long as the x is never 0. Fortunately, since we are taking a limit as x *approaches* 0, we don't care what happens *at* zero. So, we now have:



This new function is nice and friendly at x = 0, so we can just evaluate it at x = 0, and that number will be the limit! So:



Isn't that amazing- an algebraic solution to this very unfriendly limit.

The final question from class was this: "i’m not really understanding what to do if there’s a gap on the graph @ the number you’re approaching :X"

I imagine that this person is asking about whether the limit exists at a non-removable discontinuity. The answer is... no! At a non-removable discontinuity, the left hand limit will not be the same as the right hand limit, so there will be no limit at that value. If I am misunderstanding the question, I trust you'll tell me!

Since there is a quiz tomorrow, if anything I've written in this post doesn't make sense, please let me know via email or Google Voice ASAP. Any questions that I can't answer via email or Google Voice won't be on the quiz.