limit question 2 (urgent)

2011-05-21 8:02 pm

For the first 1, why is there no negative sign? Since the denominator is divided by a negative number, I wonder why is there no negative sign.

For the second 1, what's wrong with my answer? The correct answer is 1/3.

圖片參考：http://imgcld.yimg.com/8/n/HA01097219/o/701105210042713873441960.jpg

thanks

更新1:

http://farm4.static.flickr.com/3329/5741681547_e7e425d680.jpg what about the q58 (in the link)? both numerator and denominator are divided by a negative number, but why is there a negative sign? thanks

更新2:

then what are the differences between q1 and q58?-.- thanks

回答 (1)

pingshek

2011-05-21 9:13 pm

✔ 最佳答案

圖片參考：http://i1096.photobucket.com/albums/g340/pingshek/2011-5-21.png?t=1305954688

http://i1096.photobucket.com/albums/g340/pingshek/2011-5-21.png?t=1305954688

2011-05-22 01:47:42 補充：
Q.58. Look carefully.

In the first step, both the numerator and the denominator are divided by x, and thus there is no change in sign.

In the second step, it is:
[√(4x² + 5)]/x = -√[(4x² + 5)/x²] as x is negative. Because:
[√(4x² + 5)]/x = [√(4x² + 5)]/(-√x²) = -√[(4x² + 5)/x²]

2011-05-22 14:51:10 補充：
Q.1 and Q.58 are two different questions.

2011-05-22 15:16:44 補充：
I think the answer of Q.1 is wrong.

I have explained that the 1st step is right. However, in the 2nd step, there should be a negative sign before the surd, i.e.
- √[(9x² + 4x + 1)/x²]

This is because as x is negative,
[√(9x² + 4x + 1)]/x = [√(9x² + 4x + 1)]/-(√x²) = - √[(9x² + 4x + 1)/x²]

參考: miraco, miraco, miraco, miraco

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