桃子

For Q2, we try to express it in the form ka^x.

Then, we can decide the limit using the following rules.
If a<1, the limit is 0.
If a=1, the limit is k.
If a>1, the limit does not exist.

For the given expression,
k=25sqrt(5), a=5/sqrt(26)<1.

[ 本帖最後由 桃子 於 2011-4-22 11:51 PM 編輯 ]

kiyiptsang

引用:

原帖由 桃子 於 2011-4-22 11:49 PM 發表
For Q2, we try to express it in the form ka^x.

Then, we can decide the limit using the following rules.
If a1, the limit does not exist.

For the given expression,
k=25sqrt(5), a=5/sqrt(26)

咁其實我果個係咪冇咁正統??

桃子

引用:

原帖由 kiyiptsang 於 2011-4-22 11:51 PM 發表

咁其實我果個係咪冇咁正統??

That type of argument is not precise and will not be accepted in exam.
You can't just say 5.咩咩咩, you have to write keep the exact value sqrt(26).

It's really just a simple 2-step question.
lim 5^x    * 125     /   [26^(x/2) *  5^1/2]
x->inf
= 25sqrt(5) lim  [5/sqrt(26)]^x
                     x->inf
= 0 (because 5/sqrt(26)<1)

kiwakwok

For question 2, hope that it helps.

kiwakwok

For question 3, hope that it helps.

_versifier_

好似複雜左d 其實你知道(26)^(1/2)>5 你就可以因x->inf.就可以照式寫0

charityho

HOW About the following?

f(x) = 5^(x+3) / [26^(x/2) *5^0.5]
      = 5^(x+2.5) / 25^(x/2) (26/25)^x/2
      = 5^2.5 /(26/25)^x/2
lim       f(x) = 0
x->inf

桃子

That's correct.