ap calculus - limits, adlfjadsf Options

[*stephinika*]

yeah i've redone this question a few times and i bet its some stupid mistake but its driving me nuts...
anyone that can help?

find the limit of:
lim (as x approaches 4) (1/rootx - 1/2)/x-4

i keep on getting either a 0 in the numerator or denominator and its driving me crazy.
the first thing i did was simplify the numerator to (2-rootx)/2rootx then multiplied by the conjugate of 2+rootx leaving me with (4-x)/(4rootx+2x).
then i multiplied the conjugate again of 4root-2x and after some simplification i ended up with [(4rootx-2x)/4x]/x-4
i then multipied by the reciprocal to do the division but i ended up with {4rootx-2x)/4x(x-4)

now what?
(did that make any sense?)

edit.//
i need more help...haha

i'm so stuck, my brain isn't working. here:

find the limit:
lim (as x approaches 2) of (root(6-x) - 2)/(root(3-x)-1)

help?

and theres two more...trig limits.

lim (as x approaches 0) of (2sinx - sin2x)/xcosx

AND

lim (as x approaches 0) of (sinx)/(x + sinx)


ugh. my brain is hurting. this is what happens when you procrastinate and forget all you learned in class almost a weekago....

This post has been edited by stephinika: Oct 10 2005, 09:51 PM

[silver-rain]

Hmm, ok yay limits.
OK, I don't think you need to do the second conjugate. Instead, you could change the 4-x to -(x-4) and the (x-4) cancels out. Then, you get -1/4rootx + 2x. You just substitute the 4 for x and should get -1/16.

[*stephinika*]

omg ilu. haha thanks, it was driving me nuts.

closed

opened again...i need more help.

[silver-rain]

Hah, I actually had that first problem on my homework once...
OK, you multiply the top and bottom by the conjugates of both the numerator and denominator. So, then you get the limit as x approaches 2 of (6-x-4)(root3-x +1) / (3-x-1)(root6-x +2). When you simplify that, you can cancel out the 2-x and then get the limit as x approaches 2 of (root3-x +1)/(root6-x +2). Plug in 2 for x and you get 1/2.

Eh, I don't like trig limits, but I'll work on it.

edit//For the first trig problem, I keep getting 0, but I'm not sure if that's correct... Well anyways this is how I did it:
sin2x= 2sinxcosx, so you substitute that in for sin2x. You then get the limit as x approaches 0 of 2sinx-2sinxcosx/xcosx. You can factor out a 2sinx from the numerator, and get the limit as x approaches 0 of 2sinx(1-cosx)/xcosx. Pull the 2 in front of the limit and divide it up so it reads 2*limit as x approaches 0 of (sinx)/x * (1-cosx)/cosx, which can then be further divided into 2*limit as x approaches 0 of (sinx)/x * limit as x approaches 0 of (1-cosx)/cosx. The limit as x approaches 0 of sinx/x = 1. Then, you get 2*1*limit (x-->0) (1-cosx)/cosx. Plug in 0 for x, and solve and then you get 0.
Sorry if that's confusing or something...

Not so sure about the second problem though.

[*stephinika*]

linda... thanks sooo much. really. if anyone can still help on the last, i'll just leave this open for now...so yah.