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

Answer 1

William Abdalla

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

#lim_(xrarr0) (3^x-8^x) = 1-1=0# and, obviously, #lim_(xrarr0) x = 0#

So the limit has indeterminate form #0/0#.

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

#d/dx(3^x-8^x) = 3^x ln3 - 8^x ln8# and, of course #d/dx(x) = 1#

#lim_(xrarr0) (3^x-8^x)/x = lim_(xrarr0) (3^x ln3-8^x ln8)/1 = ln3 - ln8#

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Answer 2

Emily Ammerman

Remembering the remarkable limit:

#lim_(xrarr0)(a^x-1)/x=lna#

than:

#lim_(xrarr0)(3^x-8^x)/x=lim_(xrarr0)(3^x-1+1-8^x)/x=#

#=lim_(xrarr0)[(3^x-1)/x-(8^x-1)/x]=ln3-ln8=ln(3/8)#.

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Answer 3

Emilia Aguilar

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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Answer from HIX Tutor

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.