Self-studying BC. Please help with limits. Thanks.

Test Preparation AP Test Preparation
1

<p>2.3.10.a. What is wrong with the following equation?</p>

<p>(x^2 + x - 6) /(x-2) = x+3</p>

<p>2.3.10.b. In view of part (a), explain why the equation</p>

<p>lim<a href="(x%5E2%20+x%20-%206)%20/(x-2)">x->2</a> = lim<a href="x+3">x->2</a></p>

<p>2.3.16. Evaluate the limit, if it exists.</p>

<p>lim<a href="(x%5E2%20-%204x)%20/(x%5E2%20-%203x%20-%204)">x->-1</a></p>

<p>2.3.18. Evaluate the limit, if it exists.</p>

<p>lim<a href="(x%5E3%20-%201)%20/(x%5E2%20-1)">x->1</a></p>

<p>2.3.20. Evaluate the limit, if it exists.</p>

<p>lim<a href="((2%20+%20h)%5E3%20-%208)%20/h">h->0</a></p>

<p>2.3.26. Evaluate the limit, if it exists.</p>

<p>lim<a href="(1/t%20-%20(1%20/(t%5E2+t)))">t->0</a></p>

<p>2.3.28. Evaluate the limit, if it exists.</p>

<p>lim<a href="(((3%20+%20h)%5E-1)%20-%203%5E-1)%20/h">h->0</a></p>

<p>2.3.29. Evaluate the limit, if it exists.</p>

<p>lim[t->0] ((1 /(t*sqrt(1+t)) - 1/t)</p>

<p>2.3.30. Evaluate the limit, if it exists.</p>

<p>lim<a href="(sqrt(x)%20-%20x%5E2)%20/(1%20-%20sqrt(x))">x->1</a></p>

2

<p>For the first one, it probably needs a "when x!=2" As for the rest, learn to factor.</p>

3

<p>I did factor/rationalize, but it still doesn't work. After I simplified it and plug in the limit for the variable, I still get 0 in the denominator.</p>

4

<p>2.3.16</p>

<p>lim <a href="x(x-4)">x->-1</a>/(x-4)(x+1) = lim [x->-1] x/(x+1)</p>

<p>The limit does not exist. To check: the limit approaches negative infinity from the left and positive infinity from the right of -1. There is an asymptote at x = -1 so the limit DNE. </p>

<p>2.3.18</p>

<p>This one is a test of recognizing an obscure factor rule, called the difference of cubes. General form: a^3 – b^3 = (a – b)(a^2 + ab + b^2). So:</p>

<p>lim<a href="(x%5E3%20-%201)%20/(x%5E2%20-1)">x->1</a> = lim<a href="(x-1)(x%5E2+x+1)">x->1</a>/((x+1)(x-1))
= lim<a href="x%5E2+x+1">x->1</a>/(x+1) No problems now with finding the limit. Limit = 3/2</p>

<p>2.3.20 I don't feel like doing the work for this. You will have to actually cube (2+h). What should happen is the only term without an h is an 8 which will drop out allowing you to cancel that h on the bottom. Not a difficult one, just time consuming.</p>

<p>2.3.26</p>

<p>lim<a href="(1/t%20-%20(1%20/(t%5E2+t)))">t->0</a> We want one unified fraction so:</p>

<p>lim[t->0] (1/t - (1/t(t+1)) Multiply that first fraction (1/t) by another name for one, aka, (t+1)/(t+1). We have a common denominator, so we now deal with one fraction by subtracting across the top:</p>

<p>lim<a href="(t+1)-1">t->0</a>/t(t+1) The top becomes t, so it cancels out with part of the bottom leaving us with lim[t->0] 1/(t+1) = 1</p>

<p>2.3.28 The only thing you have to remember here is that when something is to the -1 power, that you just put it below 1 as in 5^-1 = 1/5</p>

<p>I might look at 29 or 30 later. These seem harder than the actual AP test but that is probably a good thing.</p>

5

<p>I don't want to discourage you. However, if you already need help on the precalculus questions, you better rethink whether or not it's a good idea to self-study BC.</p>