Limits and continuity practice test solutions PDF

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Practice problems for Limits and Continuity....

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Limits and Continuity Practice Test 1. Find lim𝑥→0 a.

2

3

6𝑥5 −8𝑥3 9𝑥3 −6𝑥5

8

b. − 9

c.

4

3

8

d. − 3

e. Nonexistent

We see that substituting in the 0 will leave us with the indeterminate form,00, so we have more work to do. We can begin by factoring a 2𝑥 3 from the numerator and a 3𝑥 3 from the denominator. This leaves us with: 2𝑥 3 (3𝑥 2 − 4) 𝑥→0 3𝑥 3 (3 − 2𝑥 2 ) lim

This allows us to cancel out the 𝑥 3 terms, giving us:

2(3𝑥 2 − 4) 𝑥→0 3(3 − 2𝑥 2 ) lim

Now, we are able to substitute in without getting the indeterminate form: 2(3 ∙ 02 − 4) 2(−4) 8 = =− 2 9 3(3 − 2 ∙ 0 ) 3(3) 2. lim𝑥→−∞(5𝑥 − 1) =

Because this is a spherical cow, the 1 will disappear. Plugging in, we get: 5 ∙ −∞ which is simply −∞.

𝑎𝑥4 +6

3. The function 𝑓 is given by 𝑓(𝑥) = 𝑥 4 +𝑏 . The figure to the right shows a portion of the graph of 𝑓. Which of the following could be the values of the constants 𝑎 and 𝑏? a. 𝑎 = −3, 𝑏 = −1 d. 𝑎 = 3, 𝑏 = −1 b. 𝑎 = 3, 𝑏 = 1 e. 𝑎 = 6, 𝑏 = −1 c. 𝑎 = 3, 𝑏 = −1

We see a vertical asymptote at 𝑥 = −1 and a horizontal asymptote at 𝑦 = −3. Recall that vertical asymptotes are manifested in the denominator of a rational function. We need our denominator to equal zero when 𝑥 = −1. Let’s look at it: 𝑥4 + 𝑏 = 0

Substitute the 𝑥 = −1 which is where we want it to actually be 0. (−1)4 + 𝑏 = 0 1+𝑏 =0 𝑏 = −1

That rules out answer choice (b). Now, to get the horizontal asymptote, in case you haven’t realized this, horizontal asymptotes are just spherical cows. You want to know where the function goes as you approach ∞ or −∞. Let’s look at the function and see what our spherical cow would be: 𝑎(∞)4 + 6 (∞)4 + 𝑏

The 6 and the 𝑏 are WAY smaller than ∞, so we can get rid of them: 𝑎(∞)4 ( ∞) 4

You should notice that the ∞s will cancel out leaving us with 𝑎. So, 𝑎 = −3 (our horizontal asymptote). Essentially, you’re solving this: 𝑎𝑥 4 + 6 = −3 𝑥→∞ 𝑥 4 + 𝑏 lim

So those are our answers: 𝑎 = −3 and 𝑏 = −1.

4. Find lim𝑥→−∞ a.

3 −2

(3𝑥−1)�𝑥 2 −4�

(2𝑥+1)2 (𝑥−1)

b.

3 2

c.

3

4

d. 1 e. ∞

We see that, because we’re taking the limit to −∞, this is a spherical cow. Therefore, we need to see whether it’s top-heavy, bottom-heavy, or equally balanced. On the top, we would end up with a 3𝑥 3 as our biggest term if we were to FOIL out. On the denominator, we would end up with a 4𝑥 3 as our biggest term if we were to FOIL out (make sure you observe that the 2 is also being squared, not just the 𝑥 ). Therefore, this function is equally balanced, and we only need to look at the coefficients to get our answer of 43. 5. The functions 𝑓 and 𝑔 are continuous. The function ℎ is given by ℎ(𝑥) = 𝑓�𝑔(𝑥)� − 𝑥. The table below gives values of the functions. Explain why there must be a value for 𝑡 for 1 < 𝑡 < 4 such that ℎ(𝑡) = −1.

They’re asking us to prove that a certain 𝑦 −value exists in an interval. This is the Intermediate Value Theorem. To prove this, we need to show that ℎ(1) and ℎ(4) surround the −1. I will use the table to evaluate at 1 and 4. Remember, because 𝑔 is on the inside, you have to get that first, and then plug that answer into 𝑓. ℎ(1) = 𝑓�𝑔(1)� − 1 = 𝑓(3) − 1 = −3 − 1 = −4 ℎ(4) = 𝑓�𝑔(4)� − 4 = 𝑓(2) − 4 = 8 − 4 = 4

So, −4 is smaller than −1 and 4 is bigger than −1. So, by the Intermediate Value Theorem, there must exist some ℎ(𝑡) = −1 in the interval [1, 4]. Note: It’s really important that you come out and specify the Intermediate Value Theorem so they know that you know what theorem you’re using.

6. Let 𝐹(𝑥) =

𝑥 2 −5𝑥−6 � 𝑥−6

, 𝑥≠6

3𝑘 + 2,

 𝑥=6

a. Find lim𝑥→6 𝐹(𝑥). Show all proper steps.

We see that the top equation is everywhere except at 𝑥 = 6. That’s what we want to use for our limit. The function is going towards 6 from both sides, it just skips the actual 6 and uses the other equation to do that one. Furthermore, because it’s the same equation for both sides, we don’t have to worry about the two one-sided limits. It’d be the same equation for both. Plugging in the 𝑥 = 6, we get:

𝑥 2 − 5𝑥 − 6 lim 𝑥→6 𝑥−6

36 − 30 − 6 0 = 6−6 0

It’s INDETERMINATE! We have more work to do… Let’s try factoring:

(𝑥 − 6)(𝑥 + 1) = lim(𝑥 + 1) 𝑥→6 𝑥→6 𝑥−6 lim

Now, we can plug in without any trouble:

=6+1= 7 b. Find the value 𝑘 such that lim𝑥→6 𝐹(𝑥) = 𝐹 (6). Show all work.

We already know from part (a) that our limit is at 7. We just need that second equation from the original piecewise function to also equal 7: 3𝑘 + 2 = 7 3𝑘 = 5 5 𝑘= 3

7. The figure shows the graph of 𝑓(𝑥). Which of the following statements are true? I. lim𝑥→1− 𝑓(𝑥) exists II. lim𝑥→1+ 𝑓(𝑥) exists III. lim𝑥→1 𝑓(𝑥) exists a. I only b. II only c. I and II only

d. I, II, and III only e. None are true

Looking at the graph, you should recognize that lim𝑥→1− 𝑓(𝑥) and lim𝑥→1+ 𝑓(𝑥) are both approaching 3. Therefore, they definitely exist, and since they are the same number, lim𝑥→1 𝑓(𝑥) also exists, meaning that I, II, and III are all true.

8. lim𝑥→5

𝑥

𝑥−5

=

Plugging in, it turns out that we get 05. This is undefined meaning our limit will be DNE, ∞, or −∞. To figure out which one, we just need to plug in test points. Since we are looking the limit going to 5, it makes sense to try 4.999 and 5.001: 4.999 + = ⇒ −∞ 4.999 − 5 − 5.001 + = ⇒ +∞ 5.001 − 5 +

Since one limit goes to ∞ and the other goes to −∞, the two one-sided limits do not agree and our limit DNE. 9. Given 𝑓(𝑥) =

6𝑥+1

√4𝑥2 +6𝑥+9

, write an equation for any horizontal asymptote(s) of 𝑓(𝑥).

Horizontal asymptotes = spherical cows! 6𝑥 + 1

𝑥→∞ √4𝑥 2

lim

=

+ 6𝑥 + 9 6(∞)

�4(∞)2 6(∞) = 2(∞) =3

So, our horizontal asymptote equation on our way to ∞ would just be 𝑦 = 3. The question definitely made it seem more complicated than that! Let’s see what happens if we plug in a −∞:

6(−∞)

�4(−∞)2 −6(∞) = �4(∞2 ) −6(∞) = 2(∞) 6 =− 2 = −3 So, we have two asymptotes at 𝑦 = 3 and 𝑦 = −3.

10. Given the function 𝑓(𝑥) =

1 1 − 𝑥+5 5

𝑥

, determine lim𝑥→0 𝑓(𝑥).

This problem is a bit of a disaster. If we try to plug in 𝑥 = 0, we see that we run into trouble. Specifically, we get: 1 1 − 0+5 5 =0 0 0

Indeterminate tells us that we need to do some simplifying. The best approach is to combine the fractions in the numerator into one fraction. That way, we can address the bigger fraction. I need those two fractions to have common denominators. I’ll 𝑥+5 multiply the first fraction by 55 and the second fraction by𝑥+5 5 𝑥+5 − 5(𝑥 + 5) 5(𝑥 + 5) 𝑥

We can now combine these two fractions:

5 − (𝑥 + 5) 5(𝑥 + 5) 𝑥

Simplifying gives us:

We can now rewrite the big fraction: The 𝑥 s cancel out giving us:

−𝑥 5(𝑥 + 5) 𝑥

−𝑥 1 ∙ 5(𝑥 + 5) 𝑥 −1 5(𝑥 + 5)

We’re finally ready to do the whole limit thing!!

1 −1 −1 =− = 𝑥→0 5(𝑥 + 5) 25 5(5) lim

11. For the function 𝑓(𝑥) shown below, find lim𝑥→0 𝑓(𝑥).

Regardless of the side you come from, this function is going down towards −∞ as you approach 0. Therefore, lim𝑥→0 𝑓(𝑥) = −∞.

12. If 𝑎 ≠ 0 and 𝑛 is a positive integer, then lim𝑥→𝑎 a. b.

1

𝑎𝑛 1 2𝑎 𝑛

c.

1

𝑎 2𝑛

𝑥 𝑛 −𝑎𝑛 𝑥 2𝑛 −𝑎 2𝑛

is e. Nonexistent

d. 0

There’s a good chance you are assuming this problem is harder than it actually is. I hope you didn’t skip it! Just like any limit, it says 𝑥 → 𝑎 , so we’re going to plug in 𝑎 wherever we see an 𝑥 : 𝑎𝑛 − 𝑎𝑛 𝑎2𝑛 − 𝑎2𝑛

This is a little snag. You see the numerator and denominator will both be zero. Since00is indeterminate, we know that we have more work to do. Here is the one legitimately tricky part of this problem. The denominator is the difference of squares. I will retype it so you can see the structure: 𝑎𝑛 − 𝑎𝑛 (𝑎 𝑛 )2 − (𝑎 𝑛 )2

So now, we can use the difference of squares to simplify the denominantor: 𝑎𝑛 − 𝑎𝑛 (𝑎𝑛 − 𝑎𝑛 )(𝑎𝑛 + 𝑎𝑛 )

See the 𝑎𝑛 − 𝑎 𝑛 terms in the numerator and denominator will cancel out? That’s pretty spectacular. We end up with: 𝑎𝑛

1 1 = 𝑛 𝑛 2𝑎 +𝑎

13. What are all the horizontal asymptotes of 𝑓(𝑥) = a. 𝑦 = 3 only b. 𝑦 = −3 only c. 𝑦 = 2 only

6+3𝑒 𝑥 3−𝑒 𝑥

in the 𝑥𝑦 −plane?

d. 𝑦 = −3 and 𝑦 = 0 e. 𝑦 = −3 and 𝑦 = 2

Again, horizontal asymptotes are just spherical cows. We’re looing at what happens as we go to ∞ and −∞. This one is a little bit different from how we would normally do it, since the 𝑥 is in the exponent, but in general, it’s the same concept. 6 + 3𝑒 𝑥 𝑥→∞ 3 − 𝑒 𝑥 lim

The 6 in the numerator and the 3 in the denominator are gone because they are just too small. 3𝑒 𝑥 𝑥→∞ −𝑒 𝑥 lim

You see that the 𝑒 𝑥 terms will also cancel out, so we basically just have: lim −3

𝑥→∞

So as we go to ∞, our function is going towards −3. We can rule out choices (a) and (c). Let’s see if it’s the same for −∞: 6 + 3𝑒 𝑥 𝑥→−∞ 3 − 𝑒 𝑥 lim

This is different! When you plug the−∞ in, you would have: 6 + 3𝑒 −∞ 3 − 𝑒 −∞

Remember, when you raise something to a negative exponent, it’s actually one over that thing: 1 3𝑒 ∞ 1 3 − 𝑒∞

6+

(it changes to a positive ∞ because I moved them to the denominator of their own little 1 fraction). 3𝑒1∞ is bottom heavy, so it goes to zero. The same thing will happen to the − 𝑒 ∞ in the denominator. So basically, what we have is: 6+0 6 = = 2. 3−0 3

Our two horizontal asymptotes are 𝑦 = −3 and 𝑦 = 2.

2 14. Given 𝑓(𝑥) = �𝑥 − 6, 𝑥 ≥ 4, find lim𝑥→4 𝑓(𝑥). 3𝑥 − 2, 𝑥 < 4

Because this is piecewise, we need to split it up and look at the one-sided limits. lim 𝑓(𝑥) = 3(4) − 2 = 10

𝑥→4 −

lim 𝑓(𝑥) = (4)2 − 6 = 10

𝑥→4 +

Therefore, our limit exists, lim𝑥→4 𝑓(𝑥) = 10.

15. The straight-line function 𝑓 is shown by the graph. Explain why there must be a value 𝑥 between 0 and 4 such that 𝑓(𝑥) = 𝜋.

This is another Intermediate Value Theorem problem because they are asking us to prove a certain 𝑦 −value occurs in a given interval. Specifically, the 𝑦 −value we are looking for is 𝜋. So, let’s see if 𝑓(0) and 𝑓(4) encapsulate the 𝜋. Looking at the graph, we see that 𝑓(0) = 1 𝑓(4) = 5

Because 𝜋 ≈ 3.14, this would certainly fall between 𝑓(0) and 𝑓(4). By the Intermediate Value Theorem, there must exist an 𝑥 such that 𝑓(𝑥) = 𝜋. Note: It’s really important that the phrase “Intermediate Value Theorem” appears in your answer somewhere. They want to know that you know what theorem to use. 16. For the function 𝑓(𝑥) graphed below, find lim𝑥→−∞ 𝑓(𝑥).

This isn’t going to have a limit. You should observe that as you go towards −∞ on the left, it bounces back and forth between 0 and −2. It will never converge on a single number, so it doesn’t have a limit.

17. The graph of 𝑓(𝑥) = √𝑥 2 + 0.0001 − 0.01 is shown in the graph to the right. Which of the following statements are true? I. lim𝑥→0 𝑓(𝑥) = 0 II. 𝑓 is continuous at 𝑥 = 0 III. 𝑓(0) is defined a. I only b. II only c. I and II only

d. I, II and III only e. None are true.

We see that the function is approaching the same value at 𝑥 = 0 from both directions. It’s actually going to 𝑦 = 0. Therefore, lim𝑥→0 𝑓(𝑥) is defined and equals 0. So, I is true. Going through our definition of continuity, 𝑓(0) is defined, lim𝑥→0 𝑓(𝑥) exists, and 𝑓(0) = lim𝑥→0 𝑓(𝑥) = 0. So, this function is continuous, and II is true.

In order for the function to be continuous, we had to establish that 𝑓(0) was defined, so III is true. I, II, and III are all true.

18. Let 𝑓(𝑥) =

2

𝑥2

and 𝑔(𝑥) = 𝑥 2 − 6. Find lim𝑥→−∞ 𝑓(𝑥) ∙ 𝑔(𝑥).

Let’s figure out 𝑓(𝑥) ∙ 𝑔(𝑥) first:

𝑓(𝑥 )𝑔(𝑥 ) =

So, we need to find the limit of that:

2 ∙ (𝑥 2 − 6) 𝑥2

2(𝑥 2 − 6) 𝑥→−∞ 𝑥2 lim

Note: You CAN combine them into one fraction, and it makes it easier to look at. You see that this is equally balanced, so our limit will be just the coefficients, 2.

𝑔(𝑥) + 𝑎, 𝑥 ≤ 0 , where 𝑎 and 𝑏 are constants 3 − 𝑏 cos 𝑥 , 𝑥 > 0 and 𝑔(𝑥) = 1 − 𝑥 2 . Show that 𝑓(𝑥) is continuous at 𝑥 = 0 if 𝑎 = 1 and 𝑏 = 1.

19. Let 𝑓(𝑥) be given by the function 𝑓(𝑥) = �

In order for 𝑓(𝑥) to be continuous, we need to show that the limit exists, the function is defined, and the limit and the function are equal. Let’s start by substituting stuff in so that it looks a little cleaner: 𝑓(𝑥) = �

Let’s start by investigating the limit:

1 − 𝑥 2 + 1, 𝑥 ≤ 0 3 − cos 𝑥 , 𝑥 > 0

lim 𝑓(𝑥) = 1 − 02 + 1 = 2

𝑥→0−

lim 3 − cos 0 = 3 − 1 = 2

𝑥→0+

Because both of our one-sided limits are equal, the limit at 𝑥 = 0 does exist. Now, we look at the function:

𝑓(0) = 1 − 02 + 1 = 2

We see that 𝑓(0) is clearly defined.

Lastly, we confirm that lim𝑥→0 𝑓(𝑥) = 𝑓(0). Indeed, lim𝑥→0 𝑓(𝑥) = 𝑓(0) = 2. So, by the definition of continuity, we have confirmed that 𝑓(𝑥) is continuous at 𝑥 = 0. 20. For the function 𝑓(𝑥) shown below, find lim𝑥→3 𝑓(𝑥).

If you will notice, the limit coming from the right is 0 and the limit coming from the left is −3. Therefore, they don’t agree and so lim𝑥→3 𝑓(𝑥) DNE....