# How do you find the limit as x goes to 0 for the function #(3^x- 8^x)/ (x)#?

I haven't figured out how to find it algebraically. But here's a non-algebraic solution:

Apply l'Hopital's Rule (it is not algebraic, it involves derivatives)

So

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Remembering the remarkable limit:

than:

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To find the limit as x approaches 0 for the function (3^x - 8^x) / x, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (ln(3) * 3^x - ln(8) * 8^x) / 1. Evaluating this expression as x approaches 0, we have (ln(3) * 3^0 - ln(8) * 8^0) / 1, which simplifies to (ln(3) - ln(8)) / 1. Therefore, the limit as x goes to 0 for the given function is (ln(3) - ln(8)).

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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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