Limits scribe (Georgette Manzur, Martha Alcocer)

Today's class, as usual, started with the following P.O.D:


*Use a graph to find the limit.

Lim 6n^2 / 3n^2 + 7n(as n approaches infinity)

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Mr. Alcantara asked "in which quadrant should we focus in order to find the answer easily", and Alejandro Covo answered that we should look the 2nd and 3rd quadrants, which was not the correct answer. Why? Because those quadrants are negative, therefore the line doesnt approach infinity.
After we graphed the function in our calculators, we all noticed that it did not reach two, so we concluded that the answer should be around that number.
Because with the graph you are not able to prove the exact answer, Mr. Alcantara teached us a new limit trick; manipulating the equation algebraically.
Mr. Alcantara first showed us the process with the P.O.D. problem:

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As you can see, after you have worked out the equation algebraically, you take the limit of the answer, which means in fact the same, but is actually much more simpler.
At that time, as shown in the bubble share slides, you're left with Lim (6)/(3 + 7/n) (as n approaches infinity). Here you can notice that as n (the only variable left) gets really big, the number is getting each time smaller because n is placed in the denominator. So because of this you can see that that whole number (7/n) will finally approach zero.
A better example of this can be seen in the P.O.D we did on Friday, which was to find lim 1/n (as n approaches infinity). By this we learned that when a variable is placed in the denominator, as that variable gets bigger the whole number gets smaller and approaches zero.
After this, Mr. Alcantara gave us another equation in order to practice and understand the new limit trick better.
Find:
Lim 6n^7+3n-5/12n^7+7n^5 (as n approaches negative infinity).
After trying by ourselves, we solved the process all together in the board:
Lim (6n^7+3n-5/12n^7+7n^5) x (1/n^7)/(1/n^7)
(6n^7/n^7+3n^2/n^7-5/n^7)/(12+7n^5/n^7)
(6+3/n^5-5/n^7)/(12+7/n^2)
At this time you can cross terms that go to zero, leaving the equation like this:
Lim 6/12 (as n approaches negative infinity) = lim 1/2

In this equation, we also solved using the fact that if n is in the denominator, the whole number will eventually become zero as n gets bigger, so that is why we crossed out all n's.
So, the answer was: 1/2
Remember, the analytical trick is better to use because you get an exact answer.
After this, Mr. Alcantara gave us several steps in order to understand the trick better. Those are as follows:
1. Only applies when x approaches positive or negative infinity.
2. You have to multiply the top and the bottom by the reciprocal (divide) of the variable raised to the highest power in the denominator.
3. Distribute.
4. Eliminate terms that go to zero.
5. Reduce the fraction.
Then he gave us three other equations to practice.
We did the first one together using the trick.
Find: Lim - ((7n^3)/(n^2-2n)) (as n approaches infinity)

First, as Mr. Alcantara said, you have to multiply the top and bottom by the reciprocal of the variable raised to the highest power in the denominator.
(DONT FORGET THAT THIS TRICK ONLY WORKS WHEN X APPROACHES INFINITY OR NEGATIVE INFINITY)

Lim -((7n^3)/(n^2-2n)) x (1/n^2)/(1/n^2)

Then you distribute:
- (7n^3)/(n^2) / (n^2/n^2)-(2n/n^2)

This equals - (7n/1-(2/n)), where you cross out 2/n because that term eventually goes to zero.

You are finally left with Lim - (7n/1) which practically equals negative infinity over one, which is the same as negative infinity. So the answer is D.N.E. (DOES NOT EXIST).