# Find d such that function tends to C as n tends to infinity

• Feb 5th 2010, 07:11 AM
harkapobi
Find d such that function tends to C as n tends to infinity
Hi, I need help on this question. I have an answer to it but i'm not sure its right. Here is goes,

Find d such that

${4^n}$/ $({\frac{1}{3^n}})^{-d}\rightarrow C$ as $n \rightarrow \infty$

Where C is a non-zero constant.

I rearranged the problem to get $(\frac{4}{3^d})^n$ and then if
d ≤ 1 then ${4^n}$/ $({\frac{1}{3^n}})^{-d}$ diverges as $n \rightarrow \infty$
d $\geq$2 then ${4^n}$/ $({\frac{1}{3^n}})^{-d}\rightarrow 0$ as as $n \rightarrow \infty$

So only value of d that converges to a non zero constant is the value of d that makes ${4^n}$/ $({\frac{1}{3^n}})^{-d}$ = 1. Is that right? so d = $\frac{ln4}{ln3}$?

Thanks
Katy :)
• Feb 5th 2010, 07:20 AM
dedust
Quote:

Originally Posted by harkapobi
I rearranged the problem to get $(\frac{4}{3^d})^n$ and then if
d ≤ 1 then ${4^n}$/ $({\frac{1}{3^n}})^{-d}$ diverges as $n \rightarrow \infty$
d $\geq$2 then ${4^n}$/ $({\frac{1}{3^n}})^{-d}\rightarrow 0$ as as $n \rightarrow \infty$

what happened when $1 < d < 2$?

Quote:

So only value of d that converges to a non zero constant is the value of d that makes ${4^n}$/ $({\frac{1}{3^n}})^{-d}$ = 1. Is that right? so d = $\frac{ln4}{ln3}$?
yes