# How do you solve: #lim_(n->oo)(ln(1+e^(2n)))/(ln(1+e^(3n)))# ? Thanks!

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To solve the limit lim_(n->oo)(ln(1+e^(2n)))/(ln(1+e^(3n))), we can use the properties of logarithms and the rules of limits.

First, we can simplify the expression by using the fact that ln(a) - ln(b) = ln(a/b).

ln(1+e^(2n)) - ln(1+e^(3n)) = ln((1+e^(2n))/(1+e^(3n))).

Next, we can apply the rule that ln(a^b) = b * ln(a).

ln((1+e^(2n))/(1+e^(3n))) = ln((1+e^(2n))/(1+e^(3n))) / ln(e^(3n)) = ln((1+e^(2n))/(e^(3n)+1)).

Now, we can take the limit as n approaches infinity.

lim_(n->oo)(ln(1+e^(2n)))/(ln(1+e^(3n))) = lim_(n->oo)(ln((1+e^(2n))/(e^(3n)+1))) / ln(e^(3n)).

Since e^(2n) and e^(3n) grow much faster than 1 as n approaches infinity, we can ignore the 1 in the numerator and denominator.

lim_(n->oo)(ln((1+e^(2n))/(e^(3n)+1))) / ln(e^(3n)) = lim_(n->oo)(ln(e^(2n))/ln(e^(3n))).

Using the property ln(a^b) = b * ln(a) again, we can simplify further.

lim_(n->oo)(ln(e^(2n))/ln(e^(3n))) = lim_(n->oo)(2n)/(3n) = 2/3.

Therefore, the solution to the given limit is 2/3.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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