Can Graphs Help Estimate Limits in Calculus Problems?

[neutron star]

Okay, so my alarm clock magically didn't go off this morning and I missed calculus (my first class missed of the year, grr!) :(

Unfortunately the teacher went over limits today and we have work due this weekend online with limits and I don't have class tomorrow. I'm having trouble figuring out how to solve these problems. Can someone help?

Homework Statement

The first one is really confusing, it is just a graph and it says Use the figure below to give an approximate value for the limit
lim f(x) (if it exists).
x$$\rightarrow$$1
http://img121.imageshack.us/img121/1862/picture4r.png

There is another problem just like this one.

Then there is another graph problem but it is different, now I'm really confused. It says to estimate the limits using the graphs.

http://img4.imageshack.us/img4/4196/picture5ju.png

I'm supposed to find:
lim ( f (x) + g (x))=
x$$\rightarrow$$5^-
lim ( f (x) + 8g(x))=
x$$\rightarrow$$5^+

Another thing I don't get is the x$$\rightarrow$$5's have + and -. What in the Universe does that mean?

The last problem is yet again different. No graph for this one. It just says to find a value of the constant k such that the limit exists for the given function.

http://img156.imageshack.us/img156/7294/picture6rn.png
http://img156.imageshack.us/img156/6981/picture7ki.png

Homework Equations

The Attempt at a Solution

I know this may seem like a lot to ask, but I really need help to understand this, and I'm not just some lazy person who missed a class because I didn't feel like getting out of bed. I have literally been studying physics most of the week and getting to bed late. Thanks ahead of time to anyone who is willing to help explain how this works.

 

[aPhilosopher]

Honestly, it would be better if you at least skimmed the chapter on limits in your book and then asked one question at a time. It's a lot of LaTex to type up if you know what I mean.

I'll be happy to help with any specific questions you have of course!

The main idea with limits is that $$\lim_{x \to a}f(x)$$ is the value that "f gets closer and closer" to as "x gets closer and closer to a" if that helps you with the first ones.

 

[neutron star]

I have skimmed through it, it does a really bad job of explaining how to do these. I'm trying to do the first problem right now. I'm guessing you have to do something with all of the open and closed circles to figure out the limit, but I'm clueless to how to go about setting that up. :(

 

[aPhilosopher]

Well if f is continuous at a then $$\lim_{x \to a}f(x) = f(a)$$ if that helps. You can just do the first few by looking at them.

 

[neutron star]

I understand but I'm supposed to find the value, not if they are continuous. What am I not seeing?

 

[aPhilosopher]

Well, if you can see that it's continuous then you can just look and see what it's value is at 1! That's the limit.

 

[neutron star]

So, what you're saying is for $$\lim_{x \to a}f(x) = f(a)$$ the approximate value either has to be a or it is not continuous right?

 

[aPhilosopher]

If you are looking at a graph, then yes because you can't have a precise value. But in general, when you are working with functions written as formulas, the value will be precise.

 

[neutron star]

I got that one wrong, I get 2 chances on every one. The answer wasn't 1 so I put na and it was wrong too. :(

 

[aPhilosopher]

the answer is 8! That's the value of f(x) at 1 ;)

Maybe you should read the chapter a little more thoroughly

 

[neutron star]

the answer is 8! That's the value of f(x) at 1 ;)

Maybe you should read the chapter a little more thoroughly

Oh my goodness, I'm even more confused now!

 

[aPhilosopher]

Well, what's the value of f(x) at 1 in the first graph?

 

[neutron star]

Why not 3? And how would that change if the open circle at 8 wasn't above 1 on the x-axis?

 

[aPhilosopher]

oh my bad man! It's been too long since I looked at a graph like that. Maybe I shouldn't be helping out but I think that's the only thing that's going to trip me up. Now I remember how to read those graphs. I guess we're both going to learn something. At least you already got the problem wrong and it wasn't my fault :P

f(x) isn't continuous at 1 because the the value 'jumps' to 3. What is the definition of a limit that they give in the book. Maybe we should work from that.

So the limit is 8. The intuitive idea of a limit is that it's the value that f(x) get's closer and closer too.

 

[neutron star]

http://img121.imageshack.us/img121/5796/picture8u.png

 

[aPhilosopher]

Oh wow! surprise! That's a real definition. They almost never do that in calc 1.

Do you understand, given the intuitive definition that I gave, why the limit is 8 though? We should cover that first. Do you see how the graph of f(x) gets closer and closer to 8 as x gets closer and closer to 1? It doesn't actually matter what the value of f(x) at 1 is or if it's even defined!

 

[neutron star]

Not really, the graph is leaving me absolutely confused. I see there is an open circle at 8 when x is 1 and a closed circle at 3.

 

[aPhilosopher]

That means that the value of f(x) at 1 is 3. The open circle means that the graph breaks there. An open circle is what they put on a graph to indicate that it looks like the value of f is where the open circle is but that it's actually somewhere else, or undefined. If you see an open circle, then if there is a solid circle somewhere else on the some vertical line, then that is the value of the function. So there is an open circle at 8 and a solid circle at 3.

I had forgotten that and so thought that the function was continuous there. At least you learned the definition of continuity though.

 

[neutron star]

What if there is a closed circle on top and an open one at the bottom?

 

[aPhilosopher]

Then the value is wherever the closed circle is. The open circle will always be on the graph though.

 

[neutron star]

It said I got another one wrong again. I put 3 for the answer. How was this wrong?

http://img121.imageshack.us/img121/6512/picture11d.png

 

[Quincy]

It said I got another one wrong again. I put 3 for the answer. How was this wrong?

http://img121.imageshack.us/img121/6512/picture11d.png

[/URL]

For a limit to exist, both of the one-sided limits have to be equal; in that graph, they are not equal, so the limit does not exist. Remember, the limit at a point has nothing to do with whether f(x) is defined at that point.

 

[aPhilosopher]

Does anyone know what the meaning of the hollow disc at (4, 7) is? is it just there to confuse?

 

[neutron star]

So they were both equal in this graph because they ended at 3?

http://img121.imageshack.us/img121/1862/picture4r.png

 

[neutron star]

Does anyone know what the meaning of the hollow disc at (4, 7) is? is it just there to confuse?

I don't know but it's creeping me out.

 

[Quincy]

Does anyone know what the meaning of the hollow disc at (4, 7) is? is it just there to confuse?

Probably.

So they were both equal in this graph because they ended at 3?

http://img121.imageshack.us/img121/1862/picture4r.png

[/URL]
In that graph, the limit as x->3 from the left is 4 and the limit as x->3 from the right is 3; 4 =/= 3, therefore the limit does not exist.

 

[neutron star]

http://img4.imageshack.us/img4/4196/picture5ju.png

http://img121.imageshack.us/img121/1396/picture12q.png

Can someone explain this to me? I know they are equal now (I think) because the lines both end at the same point on the x-axis. So they have a limit.

 

[Quincy]

http://img4.imageshack.us/img4/4196/picture5ju.png

http://img121.imageshack.us/img121/1396/picture12q.png

Can someone explain this to me? I know they are equal now (I think) because the lines both end at the same point on the x-axis. So they have a limit.

lim as x-> 5^- (f(x) + g(x)) = lim as x-> 5^- (f(x)) + lim as x-> 5^- (g(x))

lim as x-> 5^- (f(x)) = 6
lim as x-> 5^- (g(x)) = 5

So, lim as x-> 5^- (f(x) + g(x)) = 6 + 5 = 11
----------------------------------------------------------------------------------------
lim as x-> 5⁺ (f(x) + 8g(x)) = lim as x-> 5⁺ (f(x)) + lim as x-> 5⁺ (8g(x))

lim as x-> 5⁺ (f(x)) = 7
lim as x-> 5⁺ (8g(x)) = 8(3) = 24

So, lim as x-> 5⁺ (f(x) + 8g(x)) = 7 + 24 = 31

 

[neutron star]

lim as x-> 5^- (f(x) + g(x)) = lim as x-> 5^- (f(x)) + lim as x-> 5^- (g(x))

lim as x-> 5^- (f(x)) = 6
lim as x-> 5^- (g(x)) = 5

So, lim as x-> 5^- (f(x) + g(x)) = 6 + 5 = 11
----------------------------------------------------------------------------------------
lim as x-> 5⁺ (f(x) + 8g(x)) = lim as x-> 5⁺ (f(x)) + lim as x-> 5⁺ (8g(x))

lim as x-> 5⁺ (f(x)) = 7
lim as x-> 5⁺ (8g(x)) = 8(3) = 24

So, lim as x-> 5⁺ (f(x) + 8g(x)) = 7 + 24 = 31

Ok, thanks but what do the ^+ and ^- mean?

 

[Quincy]

Ok, thanks but what do the ^+ and ^- mean?

lim as x->5^- is the same as the limit as x approaches 5 from the left of the graph and lim as x->5⁺ is the same as the limit as x approaches 5 from the right of the graph. These are called one-sided limits, since you're only concerned with when x approaches a number from either the right or the left side. The two-sided limit (or just limit) concerns x approaching a number from both sides.

 

[neutron star]

Ok thanks, how do you solve this type of problem?

http://img156.imageshack.us/img156/7294/picture6rn.png
http://img156.imageshack.us/img156/6981/picture7ki.png

 

[Quincy]

Ok thanks, how do you solve this type of problem?

http://img156.imageshack.us/img156/7294/picture6rn.png
http://img156.imageshack.us/img156/6981/picture7ki.png

[/URL]

Ok, suppose k = 4.
So you have: lim as x->infinity (2^4x + 8)/(2^3x + 5). Rather than looking at the function as a whole, look at the numerator and denominator separately.
Consider the dominant term in the function. When x is really large/close to infinity, the value of 2^4x is going to be larger than the value of 2^3x, and definitely larger than the value of 8 and 5. Since 2^4x is the dominant term, only consider the limit for that function. So, the lim as x->infinity 2^4x = ? As x gets larger and larger, the value of 2^4x also gets larger and larger, so the answer is infinity.If k = 3, then there is no one dominant term. So find the lim as x->infinity (2^3x/2^3x), which is of course equal to lim as x->infinity (1), since 2^3x/2^3x cancel.

If k< 3, then again, only consider dominant terms. So we have lim as x->infinity (2^2x/2^3x) (suppose k = 2).

As x-> infinity the denominator is going to be becoming much larger than the numerator, so the value of the function is going to get smaller and smaller, and approach zero.

I know my explanation is fuzzy, I just hope I didn't confuse you...