Evaluating lim x → 1 ∫ 1 x ln 2 ⁡ ( t ) d...

Jameson Lucero

Jameson Lucero

Answered

2022-07-06

Evaluating lim x 1 1 x ln 2 ( t ) d t ( x 1 ) 3

Answer & Explanation

verzaadtwr

verzaadtwr

Expert

2022-07-07Added 17 answers

In terms of the evaluation of the limit: applying L'Hôpital's rule once forms
lim x 1 1 x ln 2 ( t ) d t ( x 1 ) 3 = lim x 1 ln 2 ( x ) 3 ( x 1 ) 2 ,
where the chain rule implies that the derivative of the denominator is
d d x [ ( x 1 ) 3 ] = 3 ( x 1 ) 2 .
From here you can apply L'Hôpital's rule two more times to find
lim x 1 ln 2 ( x ) 3 ( x 1 ) 2 = lim x 1 ln ( x ) 3 x ( x 1 ) = lim x 1 1 x ( 6 x 3 ) = 1 3 .
To answer your second question, the result follows from the fundamental theorem of calculus. Let f be a continuous, real-valued function defined on the closed interval [a,b]. Suppose F is defined for all x [ a , b ] by
F ( x ) = a g ( x ) f ( t ) d t .
Then
F ( x ) = d d x a g ( x ) f ( t ) d t = f ( g ( x ) ) g ( x ) .
Note that we could change the lower integration constant to any other positive real constant a > 0 (as the fundamental theorem of calculus would still hold) to deduce
F ( x ) = d d x a x ln 2 ( t ) d t = ln 2 ( x ) .
However, if the upper limit of integration was x 2 instead of x, then we would find
F ( x ) = d d x a x 2 ln 2 ( t ) d t = 2 x ln 2 ( x 2 ) .

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