<p>I just wanna make sure I did this limit right.</p>
<p>lim x->0
((1/2+x)-1/2)/x</p>
<p>I made this 1/x^2+2x-2</p>
<p>So I got -1/2</p>
<p>Can you just tell me if this is correct? If it's wrong, what should I do?</p>
<p>EDIT: There's one more actually, sorry.</p>
<p>lim x-->0 (2+x)^3-8/x</p>
<p>I have no clue what to do with this one</p>
<p>like omg limits are probably the easiest things ever. you friggin just plug in the number, unless its undefined, where u can use l'hopital's rule or just factor.</p>
<p>Ok, we didn't learn l'hopital because we didn't do derivatives yet as the class just started. its obviously undefined, I just need to know if I did the first one right and how to do the second one.</p>
<p>No, I just started calc yesterday, so no but the book answer is 12.</p>
<p>anyone please? if its so simple?</p>
<p>OK you're wrong logisticswizard, since the answer to the first one is 1.</p>
<p>It simplifies to...
lim x-->0 x/x</p>
<p>Since you can't use L'Hopital's yet, you must take the limit by dividing the numerator and the denominator by the variable with the highest power, which in this case is x, so basically it just reduces to 1.</p>
<p>Wait, can't you just take the 1/2+x and the -1/2 and put them in denominator and get 1/x(2+x)-2, which simplifies to what I had before? Why wouldn't that work?</p>
<p>Nvm the other one because I figured out how to get 12, I forgot the -8 is all.</p>
<p>for your second question, you can assume that the limit of a sum is the sum of the limits</p>
<p>therefore, you can split it up to
lim (x->0) (2-x)^3 - lim(x->0) 8/x</p>
<p>the first limit = 8 by substituting 0 for x</p>
<p>the second limit comes to 8/0 which is infinity
so it becomes 8 - infinity
or
-infinity</p>
<p>(i think this is right, i havent done these kinda limits since the beginning of last year haha)</p>
<p>unless you wrote it with the wrong notation - if it was really
(2-x)^(3 - 8/x) thats a different story</p>
<p>No, 12 is the answer in the back of the book. You just expand the (2+x)^3 and cancel terms to get (x^3+6x^2+12x)/x, factor out an x and get x^2+6x+12 and plug in 0 to get 12. I got that one. It's the other one, I don't know why my reduction is wrong.</p>
<p>((1/2+x)-1/2)/x
well (1/2+x)-1/2 = x. then when u divide by x, u get x/x = 1. idk why u were putting it in the denominator</p>
<p>Okay, you guys are all WRONG!!!!!I'm wrong too. The answer is -1/4. And you guys acted like I'm the only stupid one...</p>
<p>it def isnt. i double checked it on my calculator, and it is def 1. maybe you typed it in wrong, b/c lim x--> ((1/2+x)-1/2)/x = 1</p>
<p>i agree w. cutie201.</p>
<p>i think i figured it out, and its b/c u typed it in wrong. when u typed (1/2+x)-1/2 it looked like (1/2)+x-(1/2), which is 1. HOWEVER, if u had typed it 1/(2+x)-1/2, you would get lim x-->0 (1/(2+x)-1/2)/x, which = -1/4</p>
<p>No, this is the book answer. You need to multiple and get the bottom and top by 2+x and 2 and you get 2-(2+x)/x(x+2)(2) or -x/4x+2x^2, factor out an x and plug the 0 to get -1/4. You all think you're so smart.</p>
<p>no need to be so obnoxious - you typed it in wrong, and ppl were only trying to help</p>
<p>ask your math teacher</p>
<p>No, that's not necessary. I figured it out and -1/4 is correct according to the book. Believe me, that man would just confuse things up a bit.</p>
<p>well, u know what my answer to both problems is?</p>
<p>TI-89....i managed a 100 on my calc final last year with 25 minutes to spare with that baby, best benjamin i ever spent</p>
<p>Wait, that finds limits, right? How much are they because I only have an 84</p>
<p>edit
nvm, i saw the benjamin ref.</p>