Limits

It's gonna be 0
Check the steps :

http://i.imgur.com/YOyhrNF.png?1

2015-03-09 01:54:28 補充：
你take左ln兩邊,拉左個1/x落黎,再e both sides

2015-03-09 10:53:06 補充：
原本Limit裡面係y=f(x), 用property of log, ln both sides 拉個1/x落黎,再e返佢

2015-03-10 03:28:04 補充：
sorry 我睇錯左..basically唔洗用L'Hopital's
首先ln左佢
然後你見到 ln(y)=(1/x)ln(7^x-4^x)
as x approaches 0+, ln(7^x-4^x) becomes (-) infinity
as x approaches 0+, 1/x becomes a really small (+) number
---> (- inf.)/(really small positive no.) ------> even (-) inf.
系呢度你可以analytically 話 limit of呢個quotient =0

2015-03-10 03:36:07 補充：
如果你想睇得清楚D點解係0
take e of both sides and take limit
---> y=e^((1/x)ln(7^x-4^x))
頭先我地話(1/x)ln(7^x-4^x)係 (-) infinity
所以宜家係 limit as x approaches 0+ of e^((1/x)ln(7^x-4^x))
((1/x)ln(7^x-4^x) = (-) quantity ---> 1/e^((1/x)ln(7^x-4^x))
宜家你可以睇到as x approaches 0+, bottom will become (+) infinity

2015-03-10 03:37:06 補充：
limit of 1/infinity ----> 0
sorry我以前做錯左個steps
希望宜家清楚D

2015-03-11 10:37:25 補充：
Please tell me if you get it or not. Thank you!

2015-03-12 07:28:03 補充：
awwwww=.=sorry lol seems like you know everything you needed then....keep typing wrong the things i wanted to say

2015-03-12 07:32:55 補充：
If you think I am so bad on this, I apologise. Didn't mean to keep typing errors...you may ignore all of my work if you want. Sorry!

2015-03-12 07:40:29 補充：
如果有下次=.=我或者成條寫出黎放張相上黎畀你lol希望會好D==好電腦打比較易打錯~_~

chin to - step 4 is wrong

ln y = ln(7^x - 4^x)/x does not imply
y = e^{1/x) ln(7^x - 4^x)

How about this one?
http://postimg.org/image/cgd8dmihh/

可否用 L'Hospital rule?