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I think I know how to do this but I'm completely blank at the moment. Do I differentiate 'ln' and then differentiate 2^x and 3^x in the denominator?
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.
ok so that worked but when I take 2^n common i get the answer "ln(2)+ln(1.5)" but wen i take 3^n common, i get "ln(3)+ln(2/3)" and these are different answers
You';re making a mistake somewhere.
[tex] \lim_{n\rightarrow\infty} \frac{2^n}{3^n} = 0 [/tex]
Hint
[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 anything to do with differentiating y = ln(3^{x} + 2^{x})?
It's the same thing. 2/3 is under unity. Multiplying it an infinite nr of times gives 0.
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 you really want answered!May I ask, what does taking this limit have anything to do with differentiating y = ln(3^{x} + 2^{x})?