James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Problems Plus
Before you look at the solution of the following example, cover it up and first try to solve
the problem yourself.
Example Evaluate lim
x l3S x sin t
x 2 3 yx dtD.
3 t
PS The principles of problem solving
are discussed on page 71.
Another approach is to use l’Hospital’s
Rule
Problems
SOLUTION Let’s start by having a preliminary look at the ingredients of the function.
What happens to the first factor, xysx 2 3d, when x approaches 3? The numerator
approaches 3 and the denominator approaches 0, so we have
x
l ` as x l 31 and
x 2 3
x
l 2` as x l 32
x 2 3
The second factor approaches y 3 3
ssin tdyt dt, which is 0. It’s not clear what happens to
the function as a whole. (One factor is becoming large while the other is becoming
small.) So how do we proceed?
One of the principles of problem solving is recognizing something familiar. Is there a
part of the function that reminds us of something we’ve seen before? Well, the integral
y x
3
sin t
t
has x as its upper limit of integration and that type of integral occurs in Part 1 of the
Fundamental Theorem of Calculus:
2. Find the minimum value of the area of the region under the curve y − x 1 1yx from x − a
to x − a 1 1.5, for all a . 0.
425
dt
d
y x
f std dt − f sxd
dx a
This suggests that differentiation might be involved.
Once we start thinking about differentiation, the denominator sx 2 3d reminds us of
something else that should be familiar: One of the forms of the definition of the derivative
in Chapter 2 is
Fsxd 2 Fsad
F9sad − lim
x l a x 2 a
and with a − 3 this becomes
Fsxd 2 Fs3d
F9s3d − lim
x l 3 x 2 3
So what is the function F in our situation? Notice that if we define
Fsxd − y x
then Fs3d − 0. What about the factor x in the numerator? That’s just a red herring, so
let’s factor it out and put together the calculation:
3
sin t
t
y
lim
x l3S x sin t
dtD x
3
x 2 3 yx − lim x ? lim
3 t
x l3 x l3
− 3F9s3d − 3 sin 3
3
dt
sin t
dt
t
x 2 3
− 3 lim
x l3
1. If x sin x − y x2
f std dt, where f is a continuous function, find f s4d.
0
Fsxd 2 Fs3d
x 2 3
− sin 3 (FTC1) n
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