Infinite Limit (1 Viewer)

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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.
 
Last edited:

benorin

Homework Helper
1,057
7
No l'Hospital's rule needed. Since [tex]\frac{2}{3} < 1,\left( \frac{2}{3}\right) ^{n}\rightarrow 0 \mbox{ as }n\rightarrow\infty[/tex]
 

VietDao29

Homework Helper
1,417
1
opticaltempest said:
[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.
You can continue by noticing that:
[tex]\ln \left( \frac{2}{3} \right) < 0[/tex]
So as [tex]n \rightarrow +\infty[/tex], [tex]n \star \ln \left( \frac{2}{3} \right) \rightarrow - \infty[/tex], right?
So as [tex]n \rightarrow +\infty[/tex], [tex]\ln y \rightarrow - \infty[/tex]
So what's [tex]y \rightarrow ?[/tex]
-----------------
Or as benorin has pointed out:
If |a| < 1 then [tex]\lim_{n \rightarrow \infty} a ^ n = 0[/tex]
If a = 1 then [tex]\lim_{n \rightarrow \infty} a ^ n = 1[/tex]
Can you get this? :)
 

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