# Leibniz' rule

• November 4th 2009, 04:40 AM
pepsi
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)
• November 4th 2009, 04:58 AM
tonio
Quote:

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)

Bottom line, you need:

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

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

Tonio