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Math 221 Worksheet 3
Wednesday 09/04/
Group: Name:
Continuity, Limits at Infinity
Instructions. Put the first and last name of everyone in your workgroup at the top of your paper. Everyone is to do their own worksheet but only one from each group is graded with the score shared. Be sure to show your work and explain your reasoning.
- (^) x→−∞lim
2 x^2 + 1 3 x − 5
We have
x→−∞lim
2 x^2 + 1 3 x − 5 =^ x→−∞lim
x^2
2 + (^) x^12 x(3 − (^) x^5 )
= (^) x→−∞lim
−x
2 + (^) x^12 x(3 − (^5) x )
= (^) x→−∞lim
2 + (^) x^12 (3 − (^5) x )
= −
3 − 0 =^ −
NOTE: A common mistake in calculating this limit is incorrectly writing
x^2 = x. This is not true if x < 0. Since we are taking the limit as x approaches −∞, (i.e. x is becoming larger and larger in the negative direction), we have
x^2 = −x in this case.
- (^) xlim→∞
x^2 + 1 − x
We have
xlim→∞
x^2 + 1 − x = lim x→∞
x^2 + 1 − x
) (^ √x (^2) + 1 + x √ x^2 + 1 + x
= lim x→∞
x^2 + 1
− x^2 √ x^2 + 1 + x = lim x→∞^ (x
(^2) + 1) − x 2 √ x^2 + 1 + x = lim x→∞
√^1
x^2 + 1 + x = 1 +∞
NOTE: It is tempting to avoid these calculations and instead say as x approaches infinity that both
x^2 + 1 and x approach ∞ as well; hence, the limit is ∞ − ∞ = 0. But this is incorrect. In fact, ∞ − ∞ 6 = 0. To see why, consider the following limit limx→∞ (x − (x + 1)). Here again, it is tempting to reason that x and x + 1 approach ∞ as x → ∞ and thus the limit is ∞ − ∞ = 0, but upon seeing that x − (x + 1) = − 1 , the correct answer for the limit is − 1.
- (^) x→lim+∞^ x
(^2) + x 3 − x
We have
x→lim+∞^ x
(^2) + x 3 − x
= (^) x→lim+∞^ x(x^ + 1) x
x −^1
= (^) x→lim+∞
x + 1 3 x −^1 =
0 − 1 =^ −∞.
- Determine the equation for each horizontal asymptote on the graph of the following function. Your answer must be justified with limits.
f (x) = 36 + 42e
6 x 2 e^6 x^ − 46
Recall that limx→−∞ ex^ = 0 or equivalently limx→∞ e−x^ = 0. With this, we have
x→−∞lim
36 + 42e^6 x 2 e^6 x^ − 46 =
0 − 46 =^
− 46 =^ −^
and
xlim→∞^ 36 + 42e
6 x 2 e^6 x^ − 46
= lim x→∞^ e
6 x(36e− 6 x (^) + 42) e^6 x(2 − 46 e−^6 x)
= lim x→∞
36 e−^6 x^ + 42 2 − 46 e−^6 x^ =
The horizontal asymptotes are y = − 1823 and y = 21.
7.? Use the Squeeze Theorem to find (^) x→lim+∞^ cos(x
(^2) sin(x)) ex
Recall that − 1 ≤ cos(x) ≤ 1 for any real number x. Thus, we get that − 1 ≤ cos(x^2 sin(x)) ≤ 1. Dividing everything by ex, we see that − 1 ex^
≤ cos(x
(^2) sin(x)) ex^
ex^
Notice that limx→+∞ − (^) e^1 x = limx→+∞ (^) e^1 x = 0. So by the squeeze theorem, we get that limx→+∞ cos(x
(^2) sin(x)) ex^ = 0.
- Sketch a graph of the functions below, and fill in the following limits at infinity: ex^ e−x^ ln(x) − ln(x) sin(x) arctan(x) For ex^ :
For e−x^ :
For ln(x) :
For − ln(x) :
For sin(x) :
For arctan(x) :
(a) (^) x→lim+∞ ex^ = +∞
(b) (^) x→−∞lim ex^ = 0
(c) (^) x→lim+∞ e−x^ = 0
(d) (^) x→−∞lim e−x^ = +∞
(e) (^) x→lim+∞ ln(x) = +∞
(f) (^) x→−∞lim ln(x) = DN E
(g) (^) x→lim+∞ − ln(x) = −∞
(h) (^) x→−∞lim − ln(x) = DN E
(i) (^) x→lim+∞ sin(x) = DN E
(j) (^) x→−∞lim sin(x) = DN E
(k) (^) x→lim+∞ arctan(x) =
π 2
(l) (^) x→−∞lim arctan(x) = − π 2
