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Limits and Continuity Practice Test, Summaries of Calculus

Washington State University (WSU or Wazzu)Calculus

Show all proper steps. We see that the top equation is everywhere except at = 6. That's what we want to use for our limit ...

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Limits and Continuity Practice Test

1. Find lim𝑥→06𝑥5−8𝑥3

9𝑥3−6𝑥5

a. 2

3

b. −8

9

c. 4

3

d. −8

3

e. Nonexistent

We see that substituting in the 0 will leave us with the indeterminate form, 0

0, so we have more

work to do. We can begin by factoring a 2𝑥3 from the numerator and a 3𝑥3 from the

denominator. This leaves us with:

lim

𝑥→0

2𝑥3(3𝑥2−4)

3𝑥3(3−2𝑥2)

This allows us to cancel out the 𝑥3 terms, giving us:

lim

𝑥→0

2(3𝑥2−4)

3(3−2𝑥2)

Now, we are able to substitute in without getting the indeterminate form:

2(3 ∙02−4)

3(3−2∙02)=2(−4)

3(3) =−8

9

2. lim𝑥→−∞(5𝑥−1) =

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

simply −∞.

Discover Summaries of Calculus Washington State University (WSU or Wazzu)

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Limits and Continuity Practice Test

Find lim𝑥→06𝑥^

(^5) −8𝑥 3 9𝑥 3 −6𝑥 5 a. (^23) b. − (^89)

c. (^43) d. − (^83) 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 2 𝑥 3 from the numerator and a 3 𝑥 3 from the

denominator. This leaves us with:

lim 𝑥→

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

lim 𝑥→

Now, we are able to substitute in without getting the indeterminate form:

3(3) =^ −^

lim𝑥→−∞(5𝑥 − 1) =

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

simply −∞.

The function 𝑓 is given by 𝑓(𝑥) = 𝑎𝑥^

(^4) + 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 b. 𝑎 = 3, 𝑏 = 1 c. 𝑎 = 3, 𝑏 = − 1

d. 𝑎 = 3, 𝑏 = − 1 e. 𝑎 = 6, 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:

Substitute the 𝑥 = − 1 which is where we want it to actually be 0.

(−1)^4 + 𝑏 = 0

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:

𝑥→∞^ lim

𝑥 4 + 𝑏 =^ −^3

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

Let 𝐹(𝑥) = (^) �

𝑥 2 −5𝑥− 𝑥−6 ,^ 𝑥 ≠^6 3 𝑘 + 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.

lim 𝑥→

Plugging in the 𝑥 = 6, we get:

It’s INDETERMINATE! We have more work to do…

Let’s try factoring:

𝑥→6^ lim

𝑥 − 6 = lim 𝑥→6(𝑥^ + 1)

Now, we can plug in without any trouble:

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 :

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.

lim𝑥→5𝑥−5𝑥 =

Plugging in, it turns out that we get 50. 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:

+ ⇒^ +∞

Since one limit goes to ∞ and the other goes to −∞, the two one-sided limits do not agree and

our limit DNE.

Given 𝑓(𝑥) = (^) √4𝑥6𝑥+1 2 +6𝑥+9, write an equation for any horizontal asymptote(s) of 𝑓(𝑥).

Horizontal asymptotes = spherical cows!

𝑥→∞^ lim

�4(∞)^2

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 −∞:

�4(−∞)^2

�4(∞^2 )

So, we have two asymptotes at 𝑦 = 3 and 𝑦 = − 3.

If 𝑎 ≠ 0 and 𝑛 is a positive integer, then lim𝑥→𝑎 𝑥^

𝑛 (^) −𝑎 𝑛 𝑥 2𝑛^ −𝑎 2𝑛^ is a. (^) 𝑎^1 𝑛 b. (^) 2𝑎^1 𝑛

c. (^) 𝑎^1 2𝑛 d. 0

e. Nonexistent

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:

𝑎 𝑛^ − 𝑎 𝑛

𝑎 2𝑛^ − 𝑎 2𝑛

This is a little snag. You see the numerator and denominator will both be zero. Since 00 is

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:

𝑎 𝑛^ − 𝑎 𝑛

(𝑎 𝑛)^2 − (𝑎 𝑛)^2

So now, we can use the difference of squares to simplify the denominantor:

𝑎 𝑛^ − 𝑎 𝑛

(𝑎 𝑛^ − 𝑎 𝑛)(𝑎 𝑛^ + 𝑎 𝑛)

See the 𝑎 𝑛^ − 𝑎 𝑛^ terms in the numerator and denominator will cancel out? That’s pretty

spectacular. We end up with:

𝑎 𝑛^ + 𝑎 𝑛^ =^

What are all the horizontal asymptotes of 𝑓(𝑥) = 6+3𝑒^

𝑥 3−𝑒 𝑥^ in the^ 𝑥𝑦 −plane? a. 𝑦 = 3 only b. 𝑦 = − 3 only c. 𝑦 = 2 only

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.

𝑥→∞^ lim

The 6 in the numerator and the 3 in the denominator are gone because they are just too small.

𝑥→∞^ lim

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

𝑥→∞^ lim −^3

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 −∞:

𝑥→−∞^ lim

This is different! When you plug the−∞ in, you would have:

Remember, when you raise something to a negative exponent, it’s actually one over that thing:

6 + 3 𝑒^1 ∞

3 − 𝑒^1 ∞

(it changes to a positive ∞ because I moved them to the denominator of their own little

fraction). 3𝑒^1 ∞ is bottom heavy, so it goes to zero. The same thing will happen to the − 𝑒^1 ∞ in

the denominator. So basically, what we have is:

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

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.

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

Let’s figure out 𝑓(𝑥)^ ∙ 𝑔(𝑥) first:

𝑥 2 ∙^

So, we need to find the limit of that:

𝑥→−∞^ 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.

Let 𝑓(𝑥) be given by the function 𝑓(𝑥) = � 3 𝑔 − 𝑏(𝑥) + cos^ 𝑎 𝑥,^ , 𝑥 ≤𝑥 > 0^0 , where 𝑎 and 𝑏 are constants

and 𝑔(𝑥) = 1 − 𝑥 2. Show that 𝑓(𝑥) is continuous at 𝑥 = 0 if 𝑎 = 1 and 𝑏 = 1.

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) = �^1 − 𝑥^

3 − cos 𝑥 , 𝑥 > 0

Let’s start by investigating the limit:

𝑥→0^ lim−^ 𝑓(𝑥) = 1^ −^02 + 1 = 2 𝑥→0^ lim+^3 −^ cos 0 = 3^ −^ 1 = 2

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

Now, we look at the function:

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.

For the function 𝑓(𝑥) shown below, find lim𝑥→3 𝑓(𝑥).

Limits and Continuity Practice Test, Summaries of Calculus

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Limits and Continuity Practice Test

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 2 𝑥 3 from the numerator and a 3 𝑥 3 from the

denominator. This leaves us with:

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

Now, we are able to substitute in without getting the indeterminate form:

3(3) =^ −^

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

simply −∞.

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:

Substitute the 𝑥 = − 1 which is where we want it to actually be 0.

(−1)^4 + 𝑏 = 0

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:

𝑥 4 + 𝑏 =^ −^3

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

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:

It’s INDETERMINATE! We have more work to do…

Let’s try factoring:

Now, we can plug in without any trouble:

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 :

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.

Plugging in, it turns out that we get 50. 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:

+ ⇒^ +∞

Since one limit goes to ∞ and the other goes to −∞, the two one-sided limits do not agree and

our limit DNE.

Horizontal asymptotes = spherical cows!

�4(∞)^2

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 −∞:

�4(−∞)^2

�4(∞^2 )

So, we have two asymptotes at 𝑦 = 3 and 𝑦 = − 3.

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:

𝑎 𝑛^ − 𝑎 𝑛

𝑎 2𝑛^ − 𝑎 2𝑛

This is a little snag. You see the numerator and denominator will both be zero. Since 00 is

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:

𝑎 𝑛^ − 𝑎 𝑛

(𝑎 𝑛)^2 − (𝑎 𝑛)^2

So now, we can use the difference of squares to simplify the denominantor:

𝑎 𝑛^ − 𝑎 𝑛

(𝑎 𝑛^ − 𝑎 𝑛)(𝑎 𝑛^ + 𝑎 𝑛)

See the 𝑎 𝑛^ − 𝑎 𝑛^ terms in the numerator and denominator will cancel out? That’s pretty

spectacular. We end up with:

𝑎 𝑛^ + 𝑎 𝑛^ =^

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