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

[opticaltempest]

How do I evaluate this limit?

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

Is this the correct approach?

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

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

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

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 $$\frac{2}{3} < 1,\left( \frac{2}{3}\right) ^{n}\rightarrow 0 \mbox{ as }n\rightarrow\infty$$

 

[VietDao29]

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

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:
$$\ln \left( \frac{2}{3} \right) < 0$$
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? :)