Leibniz' rule

hello,

i am trying to evaluate

=lim(as x -->2) of {[1/(x-2)][integral from 2 to x of [(sint)/t]dt]}

i have tried with Leibniz' rule and now derived that

z= [integral from 2 to x of (sinx/x)dt} =>

dz/dx = (sinx)/x => z = integral of (sinx)/x

but then i am still at a loss as to how to evaluate

=lim(as x -->2) of {[1/(x-2)][integral of [(sinx)/x]dx]}



Pleasssssssseeeeeeee hlp(Bow)

Quote:

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Originally Posted by pepsi

hello,

i am trying to evaluate

=lim(as x -->2) of {[1/(x-2)][integral from 2 to x of [(sint)/t]dt]}

i have tried with Leibniz' rule and now derived that

z= [integral from 2 to x of (sinx/x)dt} =>

dz/dx = (sinx)/x => z = integral of (sinx)/x

but then i am still at a loss as to how to evaluate

=lim(as x -->2) of {[1/(x-2)][integral of [(sinx)/x]dx]}



Pleasssssssseeeeeeee hlp(Bow)

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Bottom line, you need:

$\displaystyle \lim_{x\rightarrow 2}\frac{1}{x-2}\int\limits_2^x\frac{\sin t}{t}\,dt=\lim_{x\rightarrow 2}\frac{\int\limits_2^x\frac{\sin t}{t}\,dt}{x-2}$

As both numerator and denominator are derivable in a neighborhood of 2 and we get an indeterminate of the form $\displaystyle \frac{0}{0}$ , you can try L'Hospital, remembering that $\displaystyle \int\limits_2^x\frac{\sin t}{t}\,dt=F(x)-F(2)\,,\,\,with\,\,F(t)'=\frac{\sin t}{t}$

Tonio