Evaluating \mathop {\lim }\limits_{n \to \infty } \left( {\frac{2}{3}} \right)^n

[opticaltempest]

How do I evaluate this limit?

<br /> \mathop {\lim }\limits_{n \to \infty } \left( {\frac{2}{3}} \right)^n <br />

Is this the correct approach?

<br /> {\rm{Let}} \; \; \; y = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{2}{3}} \right)^n <br />

<br /> \ln y = \mathop {\lim }\limits_{n \to \infty } \ln \left[ {\left( {\frac{2}{3}} \right)^n } \right]<br />

<br /> \ln y = \mathop {\lim }\limits_{n \to \infty } \left[ {n \cdot \ln \left( {\frac{2}{3}} \right)} \right]<br />

I am stuck at this step. I don't see a way to manipulate the limit into a
form that L'Hopital's Rule will apply. I know the limit evaluates to 0.

 

[benorin]

No l'Hospital's rule needed. Since \frac{2}{3} &lt; 1,\left( \frac{2}{3}\right) ^{n}\rightarrow 0 \mbox{ as }n\rightarrow\infty

 

[VietDao29]

<br /> \ln y = \mathop {\lim }\limits_{n \to \infty } \left[ {n \cdot \ln \left( {\frac{2}{3}} \right)} \right]<br />

I am stuck at this step. I don't see a way to manipulate the limit into a
form that L'Hopital's Rule will apply. I know the limit evaluates to 0.

You can continue by noticing that:
\ln \left( \frac{2}{3} \right) &lt; 0
So as n \rightarrow +\infty, n \star \ln \left( \frac{2}{3} \right) \rightarrow - \infty, right?
So as n \rightarrow +\infty, \ln y \rightarrow - \infty
So what's y \rightarrow ?
-----------------
Or as benorin has pointed out:
If |a| < 1 then \lim_{n \rightarrow \infty} a ^ n = 0
If a = 1 then \lim_{n \rightarrow \infty} a ^ n = 1
Can you get this? :)