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ChBjo
個人資料 | email |
posted 11-16-99 12:03 PM PT (US)
談log 秋水, 我的解法如下 lim_(x -> 1) [log x / log (2-x)] = lim_(x -> 1) [(1/x)*(log e)] / -[1/(2-x)*(log e)]
= lim_(x- > 1) (1/x) / -[1/(2-x)] = lim_(x-> 1) -[ (2-x) / (x) ] = -1 lim_( x-->0 ) log (1+ x ) / x = lim_( x-->0 ) [1/(1+ x )*(log e)]/ 1/x*(log e) = lim_(x -->0) 1/(1+x) / (1/x) = lim_(x -->0) x / (1+x) = 0 |
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wuji
個人資料 | email |
posted 11-16-99 9:38 PM PT (US)
= lim_( x-->0 ) [1/(1+ x )*(log e)]/ 1/x*(log e) Does not look right to me. -> = lim (x->0) (1/(1+x)*log(e)) / (1)
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wuji
個人資料 | email |
posted 11-16-99 9:49 PM PT (US)
Try this page for your integral needs: I think this is the reason why laptops are not allowed in exams.^_^ |
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Xiren
個人資料 | email |
posted 11-16-99 10:08 PM PT (US)
Apparently, ChBjo solved the wrong problem -- Suizette's original question was: limit_( x-->0 ) { [log(1 + x )]/ x } = ? , not limit_( x-->0 ) log[(1 + x) / x] = ? So, wuji's answer is the right one. |
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wuji
個人資料 | email |
posted 11-17-99 7:04 AM PT (US)
ChBjo did solve a wrong problem, but it was not "limit_( x-->0 ) log[(1 + x) / x] = ? " that he solved, it was "limit_( x-->0 ) log[(1 + x)] / [log x] = ?". log [(1+x) / x] is equal to log (1+x) - log (x). And if you take
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YCP
個人資料 | email |
posted 11-17-99 7:27 AM PT (US)
We can't apply L'Hopital rule for limit_(x->0) log(1+x)/log(x), since log(x) @x=0 is not 0. |
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Xiren
個人資料 | email |
posted 11-17-99 7:52 AM PT (US)
log [(1+x) / x] is equal to log (1+x) - log (x). And if you take the limit to 0, it'll actually be another interesting problem. We can't apply L'Hopital rule for limit_(x->0) log(1 x)/log(x), since log(x) @x=0 is not 0. Yep, wuji and !(Old Man) are both right! I don't think the limits exist for limit_( x-->0 ) log[(1 + x) / x] or limit_( x-->0 ) log[(1 + x)] / [log x] since log(0) is undefined. |
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ChBjo
個人資料 | email |
posted 11-17-99 8:12 AM PT (US)
I should shut my mouth out since I am brain dead these days ~~~^^ |
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suizette
個人資料 | email |
posted 11-18-99 10:58 AM PT (US)
多謝各位的回答。 為了幾道數學題，又將許久以前學的微積分重新復習了一下，
又，不知是否有這樣的規定:
then the answer will be log(e) = 2.71828../2.71828.. = 1. Am I right?? I am not really sure, cus I learned my calculus in Taiwan. 案，e= 2.718281... |
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suizette
個人資料 | email |
posted 11-18-99 11:13 AM PT (US)
log(e) = 2.71828../2.71828.. = 1. 更正…
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suizette
個人資料 | email |
posted 11-18-99 11:26 AM PT (US)
唉呀呀，又錯了…′︵‵ 2.71828..﹙ 一次方﹚ = 2.71828.... 不知用什麼來表示"次方"…
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Xiren
個人資料 | email |
posted 11-18-99 11:32 AM PT (US)
No, when the base is omitted in a logarithm, it means a logarithm to the base 10. It's also called "common logarithm", so log X = log_10 X. ln (aka. natural logarithm) is a logarithm in which the base is the irrational number e (= 2.71828 . . . ), therefore ln X = log_e X. |
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suizette
個人資料 | email |
posted 11-18-99 11:32 AM PT (US)
有勞 RS, RS' 幫我修正留言… |
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wuji
個人資料 | email |
posted 11-18-99 11:32 AM PT (US)
做微積分時，如果沒有特別聲明，一般對數函數都 是以e為底，也就是說，log x 實為 ln x,， No, I don't think so. I think the "rule" is that if the base is not explicitely stated, e.g. log x, then it is assumed to be 10. This is to say that log x is NOT ln x. |
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ChBjo
個人資料 | email |
posted 11-18-99 12:43 PM PT (US)
well, in statistics, log and ln are used as the same material........(both of them will get the same result) but not in calculus |
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ChBjo
個人資料 | email |
posted 11-18-99 1:38 PM PT (US)
突然發覺一件事, 這裡每個人對微積分的學習 都不知道比我久多少, 實在不需要班門弄斧 ^_^ |
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Xiren
個人資料 | email |
posted 11-18-99 2:17 PM PT (US)
well, in statistics, log and ln are used as the same material........(both of them will get the same result) but not in calculus I'm not exactly sure what you meant, but the results of log(x) and ln(x) do not depend on the areas where they are applied. I can only think of one case in which log(x) = ln(x) -- when x=1, otherwise the results are NOT the same. For example, log(10)=1 but ln(10) = 2.302585... (Just use a calculator if you are not sure.) log_n (a) = x is basically another way of representing n^x = a, so can you say that 10^x = (2.71828...)^x, for all x <> 0 ? |
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ChBjo
個人資料 | email |
posted 11-18-99 6:26 PM PT (US)
呵呵, 說了是算出來的結果一樣, 但log 和 ln 是在算 regression line, 並不是在算 log X= ??? or ln X= ??? 所以我今年數學課特別慘!!!! 我寧願在那裡慢慢算
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