Quick Math IIC Questions

Test Preparation SAT Subject Test Preparation
1

<p>lim (x^3) + 8 / (x^4) - 16<br>
x--> 2</p>

<p>This is a relatively simple question but how do you simplfy/factor (x^3) + 8 and (x^4) - 16 so I can cancel and plug in. </p>

<p>Also does the exam have alot of "Greatest Interger Function problems"</p>

<p>I dont understand this question where f(x) = [2x] - 4x with domain 0<x<2 (<=""> both less than and greater than equall to), then f(x) can also be written as </x<2></p>

<p>a)2x
b)-x
c)-2x
d) x^2 - 4 x
e) None</p>

<p>E is the answer here.</p>

2

<p>it’s none of the above because it simplifies to </p>

<p>lim (x^2 - 2x + 4) / (x - 2 ) (x^2 + 4)
x –> 2</p>

3

<p>um no…</p>

<p>Those are two separate problems</p>

<p>The limit question does have an answer I just want to know the trick to factoring the top and bottom.</p>

4

<p>Easy Calculus BC method:</p>

<p>lim (x^3) + 8 / (x^4) - 16 = lim 3(x^2)/(4(x^3)) = 3/8</p>

<p>(Both lims are as x –> 2)</p>

<p>If you are at least in Calc AB, know that lim f(x)/g(x) = lim f’(x)/g’(x).</p>

5

<p>You should realize that the bottom would equal 0 when you are going towards 2.
so its DNE, no need to use any BC method.</p>

<p>Or get a Ti89, put it into the calculator, and you got an answer in 5 seconds.</p>

6

<p>haha sorry. </p>

<p>x^3 + 8 factors into (x+2) (x^2 - 2x + 4) —> the trick here is that x^3 + y always factors into (x+cubedroot of y) (x^2 - cubedroot of y + cubedroot of y squared)
it’s the same with x^3 - y except the signs are different.
the x + 2s cancel out and it doesn’t have a real answer.</p>

7

<p>"If you are at least in Calc AB, know that lim f(x)/g(x) = lim f’(x)/g’(x). "</p>

<p>This formula only works if the limit is undefined (0/0 or infinity / infinity) as f(x)/g(x). So, that statement is false. Ex…lim x>2 1/2*x^0 simply = 1/2 … yet lim x>2 0 = 0 … see the contradiction?! Be careful with L’Hospitals (sp?) rule, it only works in an undefined setting and many students get tripped up on this and improperly apply the technique.</p>

<p>For those of you who are still anti-89 :slight_smile: you can directly type this into an 89 and it will spit out “0.” Just a thought to make things nicer for those who want a few bonus points.</p>

8

<p>Lol, that’s right. I remembered that a couple hours after I posted…and forgot to change it.</p>

9

<p>I want a Ti-89 O___O</p>

<p>I am also broke.</p>

<p>Alas.</p>

10

<p>Ti-89s really should be banned.</p>

<p>We ought to start a thread for all the people who got an 800 in Math II without using a Ti-89.</p>

11

<p>I own a TI-89, but I use my trusty TI-84 on all standardized tests. I do like <em>somewhat</em> of a challenge…</p>

12

<p>Thank you Crystal Pineapple for specifically answering my question. </p>

<p>Also how does TI89 help so much? Doing Limits is much faster without calculators, if you know what your doing of coarse.</p>

13

<p>Of course.</p>

14

<p>^ You can use l’hospital’s rule if the denom is undefined or 0</p>

15

<p>Wait…so I was right?</p>

16

<p>“You can use l’hospital’s rule if the denom is undefined or 0” </p>

<p>First of all, I’m wondering what you mean that a denominator is undefined. Like (3/x)/(4/(x-1) is undefined at x=1. This rule does not apply to this situation. In fact when a limit is undefined L’Hospital’s rule often won’t work if the function is bound by an asymptote (sp?) to postive and negative infinity because the limit shouldn’t exist, yet you can force it by misapplying this rule.</p>

<p>Second of all, you can **only **use L’Hospital’s rule if f(x)/g(x) is an indeterminant form.</p>

<p>that means the limits must be 0/0 or inf/inf (or negative infinity). see the link for clarification.</p>

<p>notice the “if” requirements for the “then” statement to logically follow.</p>

<p>[Limits</a> and L’Hospital’s Rule](<a href=“http://en.wikipedia.org/wiki/LHopitals_rule]Limits”>L'Hôpital's rule - Wikipedia)</p>