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- Thread starter jokerzz
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Mentallic

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[tex] \lim_{n\rightarrow\infty} \frac{\ln\left(3^n\left(1+\frac{2^n}{3^n}\right)\right)}{n} [/tex]

And now use the logarithm's properties.

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You';re making a mistake somewhere. Use the fact that:

[tex] \lim_{n\rightarrow\infty} \frac{2^n}{3^n} = 0 [/tex]

[tex] \lim_{n\rightarrow\infty} \frac{2^n}{3^n} = 0 [/tex]

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- #7

VietDao29

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Hint

[tex] \lim_{n\rightarrow\infty} \frac{\ln\left(3^n\left(1+\frac{2^n}{3^n}\right)\right)}{n} [/tex]

Err.. Sorry, but I couldn't get your point... What are you trying to do?

And now use the logarithm's properties.

When dealing with taking the derivatives of composite functions

[tex]\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}[/tex]

I'll give you an example so that you can understand how to apply the rule.

Differentiate:

--------------------

Here, we notice that, if we define another function

So, by using the Chain Rule, we have:

[tex]\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx} = \frac{d(\sin u)}{du} \times \frac{d(x ^ 2)}{dx} = \cos(u) \times 2x = 2x\cos(x ^ 2)[/tex].

Let's see if you can tackle your problem. It's pretty much the same as the example above. :)

- #8

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You';re making a mistake somewhere.

[tex] \lim_{n\rightarrow\infty} \frac{2^n}{3^n} = 0 [/tex]

It is?? How come? I took it as "(2/3)^n"

- #9

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It's the same thing. 2/3 is under unity. Multiplying it an infinite nr of times gives 0.

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VietDao29

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[tex] \lim_{n\rightarrow\infty} \frac{\ln\left(3^n\left(1+\frac{2^n}{3^n}\right)\right)}{n} [/tex]

And now use the logarithm's properties.

May I ask, what does taking this limit have

- #11

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May I ask, what does taking this limit haveanythingto do with differentiating y = ln(3^{x}+ 2^{x})?

The question I originally asked was just part of the whole question. Read my post above

- #12

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It's the same thing. 2/3 is under unity. Multiplying it an infinite nr of times gives 0.

Thanks dude for your help. I understand now

- #13

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[tex] \lim_{n\rightarrow\infty} \frac{\ln\left(2^n + 3^n \right)}{n} [/tex].

First he though of setting the fraction as a function and using l'Ho^pital's rule. I only tried to show him that he needn't know any differentiation to solve his problem.

- #14

HallsofIvy

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The OP wanted to differentiate that in order to use L'Hopital's rule to do that limit. As usual, there was a far easier way to find the limit than to use L'Hopital's rule. Another case of not asking the question youMay I ask, what does taking this limit haveanythingto do with differentiating y = ln(3^{x}+ 2^{x})?

- #15

VietDao29

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