4. Find the following limits. (You can use anything we proved in class about limits, but you
need to explain your steps!)
(a)
lim
x
→
0
sin(5
x
)
√
1 +
x

√
1

x
= lim
x
→
0
sin(5
x
)(
√
1 +
x
+
√
1

x
)
(
√
1 +
x

√
1

x
)(
√
1 +
x
+
√
1

x
)
= lim
x
→
0
sin(5
x
)(
√
1 +
x
+
√
1

x
)
2
x
= lim
x
→
0
sin(5
x
)
5
x
(
√
1 +
x
+
√
1

x
)
2
/
5
The function
(
√
1+
x
+
√
1

x
)
2
/
5
is continuous at zero so its limit is
(
√
1+0+
√
1

0)
2
/
5
= 5. The
limit of
sin(5
x
)
5
x
is the same as
sin(
x
)
x
as
x
→
0 which is 1. Using the fact that the limit
of products is the product of limits we get that
lim
x
→
0
sin(5
x
)
√
1 +
x

√
1

x
= 5
.
(b)
lim
x
→∞
2
3
x
+ 5
= lim
x
→∞
2
/x
3 + 5
/x
Since lim
x
→∞
1
/x
= 0 we have lim
x
→∞
2
/x
= 0 and lim
x
→∞
3 + 5
/x
= 3 (using the
sum and product rule of limits). Then the limit of the ratio is the ratio of the limits
(since the limit of the denominator is not zero), so lim
x
→∞
2
3
x
+5
= 0.
Another way to prove this would be by the application of the squeezing principle. If
x >
0 then
0
<
2
3
x
+ 5
<
2
3
x
.
Since 1
/x
→
0 as
x
→ ∞
, this will also be true for
2
3
1
x
(by the product rule of limits).
Another proof would be by using the
ε

δ
deﬁnition (with a large constant
c
in place
of
δ
, since we take the limit at
∞
).
2
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 Fall '08
 Staff
 Math, Calculus, Derivative, Sin, lim, Continuous function

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