Posted by sniper on August 15, 2002 at 19:36:09:
In Reply to: Oh, grum pee, I am so touched by your compliment(?) of me that I am crying a river of tears. posted by Dr. Andrew Coultos on August 06, 2002 at 10:26:29:
: : : : T. Gracken,
: : : : let me identify four of your lines in your post by
: : : : using I, II, III, and IV. My versions of your lines should be equivalent to yours.
: : : : I: lim(x->0){e^ln[e^x + x]^(2/x)}
: : : : II: lim(x->0){e^[2ln(e^x + x)/x}
: : : : III: lim(x->0){e^[2(e^x + 1)/(e^x + x)]}
: : : : IV: e^4
: : : : point 1) (I) does not lead to (II). If you move the "2" down as a multiplier on the ln expression, then (I) would lead to this:
: : : : lim(x->0){e^[2ln(e^x + x)^(1/x)}.
: : : yes. that was a typo on my part. It is difficult sometimes to check my entered text on the format provided here. sorry
: : : :
: : : : point 2) You are missing a "]" in line (II). Or you
: : : : should not have had any brackets in this line at all?
: : : yes. another typo. although I did not catch it until after posting. (unlike the previous one I just overlooked entirely)
: : : :
: : : : point 3) (II) does not lead to (III). If your line (II) had been correct, then your line (III) is wrong
: : : : because it is not the limit of the derivative seen in
: : : : (II). The derivative in line (II) is more complicated
: : : : than what you typed for (III), and it is
: : : : (e^x + x)[2x(e^x + 1)/(e^x + x) - 2ln(e^x + x)].
: : : no, I used L'Hopital's rule, not the quotient derivative. but I was not about to write every step. the result does lead (provided my earlier typos were corrected
: : : :
: : : : point 4) Although (IV), the correct answer, does follow from (III), the work of showing what value the limit is, as a whole, is wrong because (III) as already mentioned has come about by a wrong method.
: : : : (If you want, print this off and hold my feet to the fire for any of the particulars of which I discussed.)
: : : thank you. I knew you'd be back.
How's the toilet-training coming along?