lim(x→+∞)[sec √(2004 + x) – sec √x] = ?

First of all, let's see the conversion as follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Sequencelim17.jpg

In fact, we can disprove the existence of the limit through expressions (1) and (2).
For expression (1), we can find a sufficiently large x such that:

√(2004 + x) = (n + 1/2)π where n is a positive integer
√(2004 + x) ± √x ≠ 2Nπ where N is a positive integer (Both NOT equal)
(since √(2004 + x) is a contiuous function which tends to infinity as x→+∞.)
So when both conditions above are satisfied, the limit becomes infinity since the numerator part is non-zero while the denominator part is zero.
For expression (2), we can find a sufficiently large x such that:

√(2004 + x) ± √x = 2Nπ where N is a positive integer (either + or -)
√(2004 + x) ≠ (n + 1/2)π and √x ≠ (n' + 1/2)π where n and n' are positive integers .... (*)
So when both conditions above are satisfied, the limit becomes zero since the numerator part is zero while the denominator part is non-zero.
Note for (*): For cos [√(2004 + x) + √x] + cos [√(2004 + x) - √x] ≠ 0, the necessary but sufficient condition is that [√(2004 + x) + √x] ± [√(2004 + x) - √x] ≠ (2k + 1)π (BOTH NOT equal), where k is a positive integer which eventually gives rise to the condition stated above.
Therefore, by the uniqueness of limit, this limit does not exist.

答案係0
你先將佢變做1/cos，然後通分母，再用Sandwich theorem，咁就可以計到個Limit架喇！
如果都係做唔到既，我可以影相send俾你
My email : [email protected]