Page 70 - Calculus Workbook For Dummies

54 Part II: Limits and Continuity

1
o Evaluate lim x sin 2 n = 0
d
x " 0 x
Here are three ways to do this. First, common sense should tell you that this limit equals 0. limx
x " 0
1
is 0, of course, and lim sin 2 n never gets bigger than 1 or smaller than –1. You could say that,
d
x " 0 x
1
therefore, lim sin 2 n is “bounded” (bounded by –1 and 1). Then, because zero × bounded = zero,
d
x " 0 x
the limit is 0. Don’t try this logic with you calc teacher — he won’t like it.
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Second, you can use your calculator: Store something small like .1 into x and then input sin 2
x
x
into your home screen and hit enter. You should get a result of + - .05. Now store 0.01 into x
1
and use the entry button to get back to sin 2 and hit enter again. The result is + .003. Now
x
x
try .001, then .0001 (giving you + - .00035 and + .00009) and so on. It’s pretty clear — though
probably not to the satisfaction of your professor — that the limit is 0.
The third way will definitely satisfy those typically persnickety professors. You’ve got to
1
sandwich (or squeeze) your salami function, sin 2 , between two bread functions that have
x
x
identical limits as x approaches the same arrow-number it approaches in the salami function.
1 1
Because sin 2 never gets bigger than 1 or smaller than –1, sinx 2 will never get bigger than
x x
x or smaller than - x . (You need the absolute value bars, by the way, to take care of negative
values of x.) This suggests that you can use b x = - x for the bottom piece of bread and
^ h
1
t x = x as the top piece of bread. Graph b x = - x , f x = x sin 2 , and t x = x at the
^ h
^ h
^ h
^ h
x
1
same time on your graphing calculator and you can see that sin 2 is always greater than or
x
x
equal to - x and always less than or equal to x . Because lim - x j = 0 and lim x = 0 and
`
x " 0 x " 0
1 1
because sin 2 is sandwiched between them, lim x sin 2 n must also be 0.
x
d
x x " 0 x
1
2
p Evaluate lim x cos x m = 0
c
x " 0
1
2
2
For lim x 2 cos m, use b x = - x and t x = x for the bread functions. The cosine of anything
c
^ h
^ h
x " 0 x
is always between –1 and 1, so x 2 cos 1 is sandwiched between those two bread functions. (You
x
should confirm this by looking at their graphs; use the following window on your graphing cal-
culator — Radian mode, xMin = –0.15625, xMax = 0.15625, xScl = 0.05, yMin = –0.0125, yMax =
1
2
2
0.0125, yScl = 0.05.) Because lim - x i = 0 and limx = 0, lim x 2 cos m is also 0.
c
_
x " 0 x " 0 x " 0 x
2
3
q lim 5 x - x + 10 = 0
4
x
x " 3 2 x + + 3
Because the degree of the numerator is less than the degree of the denominator, this is a Case 1
problem. So the limit as x approaches infinity is 0.
4
3
r lim 3 x + 100 x + 4 = 3
4
x " - 3 8 x + 1 8
4
3
3 x + 100 x + 4
lim is a Case 2 example because the degrees of the numerator and denominator
4
x " - 3 8 x + 1
3
are both 4. The limit is thus .
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