- 135

- 0

How do I evaluate this limit?

[tex]

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

[/tex]

Is this the correct approach?

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

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.

[tex]

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

[/tex]

Is this the correct approach?

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

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.

Last edited: