Limits Help!!

[oten681]

Okay so I took AP Cal in high school and now I'm about to take it in college... Today my friend texted me asking me for help on limits since he is starting AP Cal this year and I originally thought it would be a piece of cake... Pff not so much. These problems involve absolute value so I was kind of stumped but had an idea of how to get the answer... I think. Let me upload the pics of the problems he sent me from his phone... I haven't really looked at 2 and 3 but for the first pic I believe that both conditions are true as the answer for the problem. I used x^3 as f(x) as well as x^2 and a couple linear equations as well and both parts got satisfied for problem one in each equation so I believe both must be true. Enough of me talking, here are the pics and someone please help me

Thanks in advance,

Chris

** For some reason first problem is the last pic, second is second pic and third problem is first from top.

Edit: Just looked at Prob 2 and I believe both 1 and 2 are also true just as in problem one.. For some reason this seems wrong to me

 

[mmm4444bot]

for the first pic I believe that both conditions are true The first exercise has three parts (A) (B) (C); I'm not sure which two you've referenced.

I used x^3 as f(x) as well as x^2 I'm not sure what this means.

lim [x ? 5] x^3 = 125

lim [x ? 5] x^2 = 25

Hi Chris:

I'm thinking that you need to review limits.

lim [x ? 5] f(x) = 4

This statement tells us that y approaches the value 4 as x approaches the value 5 from either side. This is like saying that the graph of function f approaches the point (5, 4).

It does not say anything about f(5). In other words, when we take a limit, as x approaches some value C, we never actually let x become C; we just let it get as close as we like.

Therefore, in the first exercise, function f might not even be defined at x = 5.

Cheers,

~ Mark

 

[oten681]

Sorry, as I stated the last pic near the bottom of the post was problem one that i was referring to

 

[mmm4444bot]

Hi Chris:

Please excuse my error; I think that I missed your edits because I was typing my response when you posted them.

Yes, both statements in the bottom pic are true because (-1)(0) = 0.

This site does not necessarily place images in the uploaded order. To position images where you want them to appear in your post, use the [Place inline] button, after uploading.

Cheers,

~ Mark

PS: Here's an example of a function for the first pic. (Double-click image, to expand.)

[attachment=0:19cz064j]piecewise function.JPG[/attachment:19cz064j]



MY EDIT: Altered domain of function p to exclude negative values of x.

 

[mmm4444bot]

For the middle exercise, consider this possible function.

f(x) = |x| - 16

What is L, if x ? 0 ?

 

[oten681]

I would assume the limit would be -16 since both sides of the graph approach -16, but since it is a sharp turn I know at that point the function isn't defined, yet I'm not sure if this means the limit doesn't exist though...

 

[mmm4444bot]

I would assume the limit would be -16 since both sides of the graph approach -16

Yes, L = -16 for part (I) because f(x) continually approaches -16 as x continually approaches 0 from either side.

since it is a sharp turn I know at that point the function isn't defined

Be careful. The derivative is not defined at x = 0. Function f is defined there; f(0) = -16.

I'm not sure if this means the limit doesn't exist though

The limit exists because f(x) approaches the same value as x approaches 0 from either side. That value is -16.

Using f(x) = |x| - 16 as an example, we see that part (I) is true.

lim [x ? 0] f(x) = -16 c = 0 and L = -16

lim [x ? 0] |f(x)| = |-16| c = 0 and |L| = 16

Graphically speaking, |f(x)| means reflecting across the x-axis that part of the graph of f(x) that lies below the x-axis.

So, the vertex reflects across the x-axis from (0, -16) to (0, 16), when we change f(x) to |f(x)|.

This is just one example for which part (I) is true. Now, think about other types of functions (cubic, exponential, logarithmic, linear, etc.) where the limit L is negative as x approaches some c.

Does reflecting the negative part of the graph in the vicinity of c across the x-axis to become |f(x)| always change the negative value of L to the positive value |L| ?

If it does, then part (I) is always true.

For part (II), consider again f(x) = |x| - 16, but start with the reflected graph for |f(x)|.

We know that the limit as x approaches 0 is 16, for |f(x)|. So, in this case, L = 16.

If we "undo" the reflection, to get the graph of f(x), then the limit changes to -16.

Does |16| = -16 ? Of course not, so part (II) is not true, in this example.

I'm not giving you rigorous proof because I'm concerned that you would not grasp it, and I'm not sure I could write such a proof, off the top of my head. My posts are intended to lead you by intuition. Familiarity with the basic types of functions and their graphs is up to you. These types of exercises are abstract; they're meant to get you thinking about all the possibilities and exceptions, when absolute values of these functions are involved. For me, the easiest way to do that is to think graphically.

Others may post a different approach.

Hmmm, I was going to say something else, but now I forget what it was. (That's what happens, when I talk too much and b(T) exceeds eight bottles.) I'm going to dinner, instead.

 

[oten681]

Thank you very much for the reply, it got me understanding limits a lot more then back when I took AP Cal. I think I'm going to take the CLEP exam for Calculus after studying a bit since I was too lazy to take the AP Exam, and you've helped me a lot more in the area of limits. Thanks again for the very detailed explanation Real quick on part 1 of that last problem, by range of f it would mean the y values of the graph contain 4 correct? And since that point might be a hole and not actually defined can you say that it would contain 4 in the range?

 

[mmm4444bot]

The "point-discontinuity" on my graph at (5, 4) means that 4 is not in the range of that function.

The range is the set of all possible y-values (i.e., all possible function outputs). That piecewise function is not defined when x = 5, so y is never 4.

HOWEVER, I goofed up, yet again today*. I intended to show an example where 4 is not in the range, but I totally forgot about the part of the parabola to the left of the y-axis. DOH!

So, in my example as originally posted, 4 was actually in the range. I'm going to edit that image now, and make my function p defined only for x ? 0. Then 4 will truly be excluded from the range of function p.

The basic idea remains unchanged. There are situations where limits exist as x approaches some value (but never reaches it) while at the same time the function does not exist at that value of x. I think that's the point the exercise is trying to drive home.

Again, just because a limit exists as x ? C does NOT guarantee that f(C) exists.

*My beer consumption in bottles b(T) is a function of the interior temperature in my house T
(measured in °Fahrenheit). Today, it was b(97).

Sorry for any confusion.

 

[Deleted member 4993]

Oh I cannot resist ...........

An infinite number of mathematicians walk into a bar. The first one goes up to the bar and orders a drink. The second one goes up and orders half a drink. The third orders 1/4 drink. The fourth orders 1/8 drink, and so on. The bartender, a little overwhelmed and worried about running out of beer, asks the mathematicians, "Hey, guys, are you sure you want to do this? Isn't that a bit much?" to which the mathematicians reply, "Oh, don't worry, we know our limits."

 

[mmm4444bot]

Limit HAIKU