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How do you evaluate the limit #lim (3^x-2^x)/x# as #x->0#?

Answer 1

Emma Aldridge

#log_e(3/2)#

#(3^x-2^x)/x=2^x((3/2)^x-1)/x = 2^x(((3/2)^(0+x)-(3/2)^0)/x)#

#lim_(x->0)(3^x-2^x)/x=lim_(x->0)2^x lim_(x->0)(((3/2)^(0+x)-(3/2)^0)/x)=1 d/(dx)(3/2)^x]_(x=0) = log_e(3/2)#

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Answer 2

Scarlett Allison

To evaluate the limit lim (3^x-2^x)/x as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (ln(3)*3^x - ln(2)*2^x)/1. Plugging in x=0, we have (ln(3) - ln(2))/1, which simplifies to ln(3/2). Therefore, the limit is ln(3/2).

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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.

How do you evaluate the limit #lim (3^x-2^x)/x# as #x-&gt;0#?

How do you evaluate the limit #lim (3^x-2^x)/x# as #x->0#?