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Calculus/indeterminate forms and improper integrals

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Expert: - 10/22/2009


Question
i have a problem i've been working on for a while now and i just can't solve it. the problem is

lim sqrt(t) - t^2
t->1 ln(t)          this is supposed to be a fraction above ln(t)

i keep getting answers like

lim  t - 2t^2
t->1 2sqrt(t)

and when i plug 1 in i get -1/2. the answer in the back is -3/2

Answer
Take f(t)/g(t) as (√t - tē)/ln(t), so f(t) = √t - tē and g(t) = ln(t).

Since both the top and the bottom go to 0, there is a theorem that says that the top and the bottom can be differentiated.

f'(t) = 1/(2√t) - 2t and g'(t) = 1/t.
Putting 1 into the derivatives gives f'(1) = 1/2 - 2 = -3/2 and g'(1) = 1.

Since f'(1)/g'(1) = (-3/2)/1 = - 3/2, that is the value of the limit.  

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