Limits Involving Infinity

It's hard to follow up the Sandwich Thm...but I'll try. :)  HW help video at bottom!

Limits involving infinity really revolve around finding horizontal asymptotes and vertical asymptotes.

If you are trying to find a limit as "x" goes to infinity (or negative infinity), you are lookign for horizontal aysmpotes.  The key?  Look at the powers:
1. Degree on bottom is bigger:  Limit  = 0

Because the denominator's largest power is larger than the numerator's largest power, the denominator's highest powered term takes this expression over as x approaches inifinity. The denominator becomes larger much quicker than the denominator, therefore the limit of this expression must be zero. Because zero is a constant, this means that there is a horizontal asymptote at .

2. Degree on top = degree on bottom:  Limit = A/B (where A = leading coefficient on top, B = leading coefficient on bottom)

In this limit problem, the highest power term in the numerator is  and the highest power term in the denominator is . Since these are the highest power terms, they dominate the limit problem, and we can ignore the other terms in determining the limit. We see that the limit is the quotient of the coefficients of the highest power terms.

3. Degree on top is bigger: Divide everything by biggest power on the bottom, and let x go to infinity of negative infinity and see where your function goes.

Because the numerator's largest power is larger than the denominator's largest power, the numerator's highest powered term takes this expression over as x approaches infinity. Therefore the limit of this type of expression must be positive infinity.

The Sandwich (or Squeeze) Theorem

The Sandwich Thm is a very creative way to find limits of functions that are down right impossible to solve using algebriac techniques.  Many times, sandwich thm problems involve trig functions...although they don't have to.  This is because we can easily "start" the problem be sandwiching sine or cosine between 1 and -1...as you no doubt recall from trig. :)

As a motivation let us consider the function

When x get closer to 0, the function fails to have a limit. So we are not able to use the basic properties discussed in the previous pages. But we know that this function is bounded below by -1 and above by 1, i.e.

for any real number x. Since , we get

Hence when x get closer to 0, x² and -x² become very small in magnitude. Therefore any number in between will also be very small in magnitude. In other words, we have

 

 

 

Enjoy!

Intro to Limits

Finally...a little CALCULUS!  I know you're excited...

So what are limits?  If you remember ONE thing from class today...I hope it's that limits are what functions are APPROACHING.  NOT what they are equal to.  Keep that in mind...especially as we start to explain the deeper meaning of limits during the next few classes.

I've included a few HW problems in the video below...(also use the HW hint sheet I gave you!)

Not that we would ever HAVE to do a limit problem like this...but the example below really illustrates the idea that we are interested in what number is being APPROACHED!

Numerical Approach to Limits

Example 1: Let f(x) = 2 x + 2 and compute f(x) as x takes values closer to 1. We first consider values of x approaching 1 from the left (x < 1).

x	f(x)
0.5	3
0.8	3.6
0.9	3.8
0.95	3.9
0.99	3.98
0.999	3.998
0.9999	3.9998
0.99999	3.99998

We now consider x approaching 1 from the right (x > 1). 

x	f(x)
1.5	5
1.2	4.4
1.1	4.2
1.05	4.1
1.01	4.02
1.001	4.002
1.0001	4.0002
1.00001	4.00002

In both cases as x approaches 1, f(x) approaches 4. Intuitively, we say that lim_x→1 f(x) = 4.

NOTE: We are talking about the values that f(x) takes when x gets closer to 1 and not f(1). In fact we may talk about the limit of f(x) as x approaches a even when f(a) is undefined.

Are You Ready For Calculus? (Helpful Hints)

So your calculus teacher gave you a big packet...stuff you "should" know...but probably forgot.  Here are a few helpful reminders:

I've included a few hints in the attached video below (you may have to pause the video...it goes a bit quick!)  

You might find the notes below to be helpful as well.

Multiplying: take your time and multiply correctly, when taking a power to a power...you multiple the powers, you are NOT solving for for anything here

Factoring:  recall how to factor, don't forget the diff. of two cubes and sum of two cubes formulas

a³ + b³ = (a + b)(a² – ab + b²) a³ – b³ = (a – b)(a² + ab + b²)

How To Ace McHale's Calculus Class

1.        Sign up for calculus.

2.       Pay attention.  Easier said than done.  I’ve spent years refining techniques to make your attention wander and your mind dip into numbness.  (not true)  If you are engaged during the 81 minutes of class, you will find yourself spending much less time out of class trying figure out what the heck I did in class. 

3.       Do your homework.  Quite frankly, most of the learning takes place when you sit in the quiet of your bedroom, library, or jacuzzi, and do your HW problems.  If you understand how to do your HW problems…life should be good for you in calculus land.

a.       Do HW set assigned in chapter outline

b.      Use examples we have done in class (there will be lots)

c.       Feel free to use your book

4.      Get help.  When you are stuck on a problem, do not:

a.       Bang your head against the wall

b.      Decide that a couple hours of Playstation will help

c.       Give up in disgust, and resign yourself to a career in Slurpee Sales

Instead, get help.  From:                               

·         Mr. McHale. That’s me.

·         Your friends.

·         Other math teachers.

·         The book.

·         Your notes.

·         The world wide web

·         Mr. McHale’s Calculus Blog (bhsmchalecalc.blogspot.com)

5.       Know examples.  Calculus is usually presented as a bunch of rules with occasional examples to illustrate them.  In this class…we will do LOTS of examples.  Pay attention to them.  You may have the idea that these are randomly picked from an ocean of examples.  However, the collection of examples is more like a small pond, so you should be particularly interested in the ones that I present to you.  Mathematicians recycle good problems the same way comedians recycle jokes.

6.      Study.  Studying will be unattractive to some of you, unfamiliar to others, but the truth is that it cannot be avoided in the end.  The key to success is to study effectively and in a way that is fun…do not avoid studying.  If you cannot come to grips with this, you might consider a career in politics or some field that does not require mathematics or higher level thought.

7.       Avoid the dark side.  Almost without exception, cheating does not lead to higher grades.  You might be able to squeak through one problem here or there, but pretty soon you will be in the middle of material that depends on material previously covered.  Cheating is high risk and no reward.  Don’t do it.

The Definite Integral

Remember the difference between the indefinite integral and the definite integral:
Indefinite

• will have a "+ C"
• answer will be a function
• no upper/lower bound on integral symbol

Definite

• no "+ C"
• answer will be a NUMBER
• it has upper/lower bounds on integral symbol

Steps to solving:

1. Find the integral

2. Plug in upper bound

3. Plug in lower bound

4. Subtract step #3 from step #2

Website with worked examples and good explanations!

Remember what we are REALLY doing...calculating definite integrals is the "easy" way to add up all those infinite rectangles and get the TRUE area under a curve.

Change of Variable...the Dirty Integral

As we attempt to undo chain rule-esque intergrals remember the following:

1. Pick a good "u".  It's is usually the "inside stuff"...but doesn't have to be. 
2. Find "du/dx"...take the derivative of u.
3. Your "dx" and anything else left over in the intergral should show up once you find "du/dx".  Sub that into your intergral.  You've now successfully changed your variables from "x" to "u".  Hopefully, the new intergral is simplier.  Take the integral.
4. Resub to get "x" in your answer.

Easy as pi... :)

Differential Equations

Today we discussed solving equations involving derivatives and anti-derivatives.  Most importantly...we use some given information to solve for the dreaded "arbitrary constant"...

1. You will be given a derivative or 2nd derivative (f'(x) or f''(x))...use techniques of integration to find either f(x) or f'(x)...there will be a "+C" in your answer.
2. Use the information given to you to solve for C.
3. Repeat if necessary...

Indefinite Integrals

And so we finally learn how to "undo" the derivative.  Remember, reversing the power requires two steps:

1. Raise the power by 1.
2. Divide the coefficient by the new power.

That's it.  :)
And don't forget your 6 trig integrals:

1.)

2.)

3.)

4.)

5.)

6.)

Curve Sketching

Some Important Reminders Regarding Curve Sketching

(I will post solutions to both HW problems tomorrow night)

Increasing and Decreasing Functions

The derivative of a function can tell us where the function is increasing and where it is decreasing. If

a) f'(x) > 0 on an interval I, the function is increasing on I.

b) f'(x) < 0 on an interval I, the function is decreasing on I.

The intervals of increase and decrease will occur between points where f'(x) = 0 or f'(x) is undefined. However, these points are not necessarily critical numbers because we include x even if it is not in the domain of f. We simply want to find the intervals of increase and decrease around x, even if the function is not defined at that point.

The graph to the right illustrates this theorem. From A to B, the slope of the tangent lines are all negative, so the derivative, f'(x) is negative from A to B. The theorem above states that the function is decreasing from A to B. The graph shows that the values of the function are decreasing between A and B. Similarly, the function is also decreasing between C and D. From B to C however, the slopes of the tangent lines are positive. Therefore, the derivative is positive from B to C. The graph shows that the values of the function are increasing between B and C.

Concavity and Points of Inflection

A graph is called concave upward (CU) on an interval I, if the graph of the function lies above all of the tangent lines on I. A graph is called concave downward (CD) on an interval I, if the graph of the function lies below all of the tangent lines on I.

The second derivative of a function can tell us whether a function is concave upward or concave downward. If

a) f''(x) > 0 for all x in an interval I, the graph is concave upward on I.

b) f''(x) < 0 for all x in an interval I, the graph is concave downward on I.

The intervals of concavity will occur between points where f''(x) = 0 or f''(x) is undefined. We test the concavity around these points even if they are not included in the domain of f.

The graphs below illustrate the different forms of concavity. Remember that there are two ways in which a graph can be concave upward or concave downward. Graphs A and C illustrate the types of concavity when the function is increasing on the interval, while graphs B and D illustrate concavity when the function is decreasing on the interval. These 4 graphs cover every different form of concavity.

A point P on a curve is called a point of inflection if the function is continuous at that point and either

a) the function changes from CU to CD at P
b) the function changes from CD to CU at P

Points of inflection may occur at points where f''(x) = 0 or f''(x) is undefined, where x is in the domain of f. We must test the concavity around these points to determine whether they are points of inflection.

The graph to the right illustrates a curve with a point of inflection.

The 2nd Derivative Test

The 2nd Derivative Test...so elegant it makes the 1st derivative test look like the ugly step-sister.  If you got one thing from the notes today, hopefully it was the connection between concavity and the 2nd derivative.

Recall:

• 2nd Derivative > 0, concave up
• 2nd Derivative < 0, concave down
• 2nd Derivative = 0 or DNE, point of inflection

In order to use the 2nd Derivative Test to prove points are indeed extrema you must do the following in you HW:

1. graph (use your graphing calculator)
2. find derivative, set equal to zero to locate critical points
3. find 2nd derivative & plug critical points into 2nd derivative
• If positive...then it's a minimum
• If negative...then it's a maximum
• If it's zero...use 1st derivative test (or see if it's a POI)
4. find increasing/decreasing intervals (use critical points to help identify intervals)
5. find concave up/concave down intervals (use POIs to help identify intervals)

Example 1: Find any local extrema of f(x) = x⁴ − 8 x² using the Second Derivative Test.

f′(x) = 0 at x = −2, 0, and 2. Because f″(x) = 12 x² −16, you find that f″(−2) = 32 > 0, and f has a local minimum at (−2,−16); f″(2) = 32 > 0, and f has local maximum at (0,0); and f″(2) = 32 > 0, and f has a local minimum (2,−16).

Example 2: Find any local extrema of f(x) = sin x + cos x on [0,2π] using the Second Derivative Test.

f′(x) = 0 at x = π/4 and 5π/4. Because f″(x) = −sin x −cos x, you find that and f has a local maximum at . Also, . and f has a local minimum at .

The 1st Derivative Test

The 1st Derivative Test is simple way to prove that critical points are extrema (or are not extrema in some cases).  Make sure you do the following:

1. Find critical points
2. Test points on both sides (using the 1st derivative)
3. Compare the slopes (+ or -)...if it goes - to +, it's a min..if it goes + to -, it's a max.  If it doesn't change...then it's NOT an extrema!

Don't forget to include the increasing and decreasing intervals on your HW.  Use your graphing calculator to help with this.  

Curve sketching is a little trickier.  The following will be helpful:

• f(3) = 4 means graph the point (3,4)
• f'(x) = 0 means you have a max or min
• f'(x) DNE means you have a cusp, corner, or asymptote
• f'(x) >0 means increasing and f'(x)<0 means decreasing

See video for problem #11 and #36.

The Mean Value Theorem

Rolle's Thm on a slant!  Don't forget, when solving MVT problems, you need to show three things:

1. That f(x) is cont. on [a,b]
2. That f(x) is differentiable on (a,b) 
3. That there is AT LEAST one point between a and b, such that the derivative is equal to the slope between (a, f(a)) and (b, f(b)).

Hint:  The last problem asks you to show that the car MUST have been going that fast at least once...you can do that using the MVT.  

Two MVT Problems

Rolle's Theorem

Good ole' Michael Rolle...do you know that Rolle once argued AGAINST the use of calculus.  In fact, he even wrote a critique of calculus that was NOT very flattering.  Luckily, he came around and gave calculus one of its most useful theorems: Rolle's Theorem

While doing your Rolle's Thm HW (pg. 267, 3-8) remember you must show THREE things:

1. That the endpoints are EQUAL.  Plug your left and right hand endpoints into the original f(x) to show this.
2. You must show (or at least say) that f(x) is continuous and differentiable on the interval between [a,b].
3. Find the derivative,set equal to zero, and verify that that point exists and is in between a and b.  (if a and b are your endpoints)

Rolle's Theorem & Example

Check out video below for help on a couple of HW problems and a sweet quote from Michael Rolle himself (thanks Grant):

Critical Point Quiz

Your quiz over critical points will cover:

1. Identify extrema on a graph (find extrema over different intervals...like #3 and 4 from the HW).  Remember, open intervals can NOT be considered extrema and constant functions are considered BOTH max and mins.
2. Finding critical points.  You should be able to find places that the derivative equals 0 and DNE.  Check your critical points to make sure that they are actually in the domain of the original function. If your derivative is too hard to set equal to zero and solve, graph the derivative and look at its zeros.
3. Given a a function, you should be able to use your calculator and locate extrema over given intervals.

Critical Points (Day 2)

Important things to take from today:

1. You can not have a max/min on an open interval.
2. You can use your calculator to determine extrema (HW problems 8-10 and 50, 51)
3. Finding critical #s can be difficult...make sure you compare your critical #s with the domain of the original function (HW problems 19-29 odd)

(See video for HW hints...you will have to pause it at times)

Critical Points (Day 1)

Critical points occur at two places:

1. any where g'(x) = 0 (set derivative equal to zero)
2. any where g'(x) DNE (so look for things that would make denominator zero, negative under radical, etc)

If your derivative ends up being a rational expression (fraction with variables on top and bottom), you can set TOP = 0 to satisfy 1st condition and bottom equal to 0 to satisfy 2nd condition.

Remember:  Extrema MUST occur at critical points...but critical points do NOT necessary have to produce extrema!