Limits.pdf


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Claim:

lim

x→1

x3 + x − 2
is indeed 4.
x−1

We have
x3 + x − 2
(x − 1)(x2 + x + 2)
=
= x2 + x + 2 for x 6= 1
x−1
x−1
So
lim

x→1

x3 + x − 2
= lim x2 + x + 2 = 4.
x→1
x−1

Example 4.15. Find lim

x→∞

x2 + 1
.
3x2 + x + 7

x2 + 1
=
3x2 + x + 7

1+
3+

1
x2

1
7
+ 2
x x

−→

1
1+0
= .
3+0+0
3

Example 4.16. Let

f (x) =



x2

if x 6= 0,


1

if x = 0.

What is lim f (x) ? Answer: 0, because as we approach x = 0 from above or
x→0

below the function gets arbitrarily close to 0. (It doesn’t matter that this is
not the value of the function there.)

Example 4.17. Consider f (x) = x sin x1 . From its graph (see Figure 2),

it looks like lim x sin x1 = 0. We will see a proof soon.
x→0

8