dilloncyh

1. Why the limit of f(x) （where f(x) = 3^2x ) as x approaches negative infinity is 0? I know the syllabus includes the graph of exponential function in the cor part, and I can understand this when I try to substitute values of x into it, but is there any genral rule or properties related to this?

2.  Why is there a negative sign in the 2nd step?

3. Same as above. Why do need to add a negativ sign?

thanks

手槍仔仔

1.


get it?

doraekit

因為tends to負無限
所以全部都要加負號

夏娜醬大好

1. lim[x→ -Inf] 3^(3x) = lim[y→ +Inf] 1/ [3^(3y)] = 0

2. lim[x→ -Inf] sqrt(x^2) = -x
    你想一想, sqrt(x^2)一定是positive, 但x是→ -Inf 的, 所以應該是 -x 才對

桃子

If a>1,
a^x→inf as x→inf
a^x→0 as x→ -inf

If 0<a<1,
a^x→0 as x→inf
a^x→inf as x→ -inf

If a<0, then the limit does not exist.

deiruty

For Q 1
lim x→ -inf (3^x)
you may deem it as lim x → inf  (1/3^x)
For Q2
x is -inf
you have to add a negative sign to it so that it can be x=((x)^2)^1/2 and x is a negative number