Limits and Continuity

Summary

This video quiz guides viewers through solving a series of limit problems, focusing on techniques like direct substitution, factoring, and applying trigonometric identities. The instructor explains step-by-step solutions for each question, such as calculating limits involving complex fractions, factoring numerators and denominators, and using the squeeze theorem. The video also covers the intermediate value theorem and finding the horizontal asymptote of a function. Each problem is presented clearly, and viewers are encouraged to pause the video to attempt solving the problems before reviewing the solutions.

Takeaways

• 📊 The video emphasizes pausing to solve each problem before viewing the solution.
• 🔢 Direct substitution is used when the denominator does not approach zero, as demonstrated in problem 1.
• 🧮 Factorization is a key method in solving limits when direct substitution results in an indeterminate form, such as in problem 2.
• ✖️ Canceling out similar terms is often necessary after factoring, as shown in problem 3.
• ➕ Using conjugates is important for rational functions with square roots, as illustrated in problem 4.
• 🔄 For indeterminate forms (0/0), it’s essential to evaluate both the left and right limits to determine if the limit exists, as in problem 5.
• 🧠 Trigonometric functions can be simplified using basic trigonometric identities like sine and cosine, as demonstrated in problem 6.
• ⬆️ Horizontal asymptotes are identified by focusing on the behavior of functions as x approaches infinity, as shown in problem 7.
• 📈 The squeeze theorem helps in determining the limit of oscillating functions, such as sine of 1/x in problem 8.
• 📉 The intermediate value theorem can verify continuous functions on a closed interval and help find specific values of c, as seen in problem 9.

Q & A

What is the method used to solve the first limit problem?

Direct substitution is used because plugging in the value of 2 into the denominator does not result in zero, allowing for simple evaluation.

How do you solve a limit problem when direct substitution leads to a zero in the denominator?

When direct substitution results in a zero in the denominator, factoring the numerator and denominator is a useful technique. Cancel out common factors before evaluating the limit.

What is the common denominator used in problem 3 to simplify the complex fraction?

The common denominator used is 4x, which is distributed across both the numerator and the denominator to simplify the fraction.

How do you handle a rational function with a square root in the numerator in problem 4?

Multiply the numerator and denominator by the conjugate of the numerator to eliminate the square root. This simplifies the limit for evaluation.

Why does the limit in problem 5 not exist?

The left-hand and right-hand limits as x approaches 7 do not match, leading to the conclusion that the overall limit does not exist.

How is tangent rewritten in the solution for problem 6?

Tangent is rewritten as sine divided by cosine. Additional adjustments are made to have appropriate terms for limit evaluation.

What happens to smaller terms when x approaches infinity in problem 7?

Smaller terms like 5x and constants are insignificant compared to terms like 8x² and 2x² as x approaches infinity, simplifying the limit.

What does the Squeeze Theorem state, as used in problem 8?

The Squeeze Theorem states that if a function f(x) is squeezed between two other functions h(x) and g(x) with the same limit at a point, then f(x) must have that same limit as well.

How is the intermediate value theorem (IVT) applied in problem 9?

The IVT is applied by confirming that f(x) is continuous on the interval and that a value of c exists such that f(c) equals the target value within that interval.

What steps are followed to find the value of c that makes a function continuous in problem 10?

The two parts of the piecewise function are set equal to each other at x = 2, and then algebraic manipulation is used to solve for c.

Outlines

00:00

🧑‍🏫 Direct Substitution in Limits - Solving Example 1

This paragraph explains how to use direct substitution to solve a limit problem. The example demonstrates substituting x = 2 into a rational function and simplifying it to find the answer. After evaluating the expression, the correct solution is found to be 6, corresponding to answer choice B.

05:01

🧮 Factoring to Solve Limits - Example 2

The second paragraph focuses on solving a limit where direct substitution leads to an indeterminate form. The process involves factoring both the numerator and denominator, canceling terms, and then evaluating the limit. The result is 4/3, corresponding to answer choice D.

10:04

📐 Complex Fractions and Limits - Example 3

This paragraph covers solving a limit involving complex fractions. The approach is to multiply by a common denominator to simplify the fraction. After simplifying and using direct substitution, the final answer is -1/16, corresponding to answer choice C.

15:08

🧮 Conjugates in Limits - Example 4

Here, the method of multiplying by the conjugate is used to simplify a rational function with a square root in the numerator. After canceling terms and substituting x = 16, the final answer is 1/8, corresponding to answer choice A.

🚦 Left and Right-Hand Limits - Example 5

This paragraph discusses how to find one-sided limits and determine whether the overall limit exists. By checking the left-hand and right-hand limits around x = 7, it is concluded that the limit does not exist, leading to answer choice E.

🔢 Trigonometric Limits - Example 6

The sixth paragraph solves a trigonometric limit by rewriting the tangent function in terms of sine and cosine, manipulating terms, and using substitution. The final result is 3/5, matching answer choice B.

🏞 Horizontal Asymptotes Using Limits - Example 7

This example explains how to find the horizontal asymptote of a rational function by evaluating the limit as x approaches infinity. By simplifying the function, the horizontal asymptote is determined to be y = 4, which is answer choice E.

📊 Using the Squeeze Theorem - Example 8

This section applies the squeeze theorem to determine the limit of x * sin(1/x) as x approaches 0. Since both bounds of the function tend to 0, the limit is also 0, corresponding to answer choice B.

🔗 Intermediate Value Theorem - Example 9

This paragraph verifies the application of the Intermediate Value Theorem on a given interval. After showing that the function is continuous and f(a) <= 0 <= f(b), the value of c within the interval is found to be 1, which matches answer choice D.

✏️ Continuity and Solving for C - Example 10

The final paragraph discusses finding the value of c to make a piecewise function continuous at x = 2. After setting the two parts of the function equal and solving for c, the value is determined to be 3, which corresponds to answer choice C.

Keywords

💡Limit

A limit is a fundamental concept in calculus that describes the value that a function approaches as the input (or variable) approaches a certain point. In the video, limits are used to solve various problems, such as finding the value of a function as it nears a specific point, which is crucial for understanding the behavior of functions, especially when direct substitution is not possible.

💡Direct Substitution

Direct substitution is a method in calculus where you directly replace the variable in a function with a specific value to find the limit. In the video, this technique is used when evaluating limits where substitution doesn't result in an undefined expression, like a zero denominator. It's a straightforward way to compute limits when conditions allow.

💡Factoring

Factoring is a mathematical process of breaking down an equation into simpler components (factors) that when multiplied together give the original expression. The video frequently employs factoring to simplify complex limits, particularly when the direct substitution leads to indeterminate forms like zero over zero.

💡Indeterminate Form

An indeterminate form occurs in calculus when the limit of a function takes an uncertain form, like 0/0 or ∞/∞. The video discusses how to handle such cases using techniques like factoring or conjugates, ensuring the limit can be evaluated despite the initial undefined form.

💡Conjugate

A conjugate in mathematics typically refers to changing the sign between two terms in a binomial expression. In the video, multiplying by the conjugate is used to simplify expressions involving square roots, which helps in finding limits that would otherwise be challenging due to the presence of radicals.

💡Intermediate Value Theorem (IVT)

The Intermediate Value Theorem is a principle in calculus stating that if a continuous function has values of opposite signs at two points, then it must cross zero at some point between them. In the video, this theorem is used to demonstrate that a function's limit exists within a given interval, helping to find specific values of 'c' where the function's value is guaranteed to be zero.

💡Squeeze Theorem

The Squeeze Theorem is a method used in calculus to find the limit of a function by 'squeezing' it between two other functions that have the same limit at a particular point. In the video, the Squeeze Theorem is used to evaluate limits where direct evaluation is complex, particularly in oscillating functions like sine.

💡Horizontal Asymptote

A horizontal asymptote refers to the value that a function approaches as the input grows infinitely large or small. In the video, the concept is applied to determine the long-term behavior of rational functions, helping to find the limit of functions as x approaches infinity.

💡Continuous Function

A continuous function is one that is uninterrupted or unbroken for every value in its domain. The video discusses the importance of continuity in applying the Intermediate Value Theorem and finding limits, particularly when ensuring a function behaves predictably without any sudden jumps or gaps.

💡Trigonometric Limits

Trigonometric limits involve evaluating the limits of functions that include trigonometric expressions, such as sine, cosine, and tangent. The video demonstrates how these limits can often be simplified using known trigonometric identities and limits, such as the limit of sin(x)/x as x approaches zero.

Highlights

Introduction to video quiz, instructing viewers to pause and work on each problem before revealing the solution.

Problem 1 demonstrates direct substitution to calculate the limit using basic algebraic operations.

In Problem 2, the solution involves factoring both the numerator and denominator to simplify the expression before finding the limit.

Problem 3 covers the method of dealing with complex fractions by multiplying both the numerator and denominator by the common denominator.

In Problem 4, the concept of multiplying by the conjugate of the numerator to simplify a rational function with a square root is explained.

Problem 5 highlights the evaluation of a limit with an indeterminate form and explains how left-sided and right-sided limits affect the result.

Problem 6 introduces trigonometric functions in limits, simplifying tangent into sine and cosine for further evaluation.

For Problem 7, the horizontal asymptote of a rational function is calculated by focusing on the dominant terms as x approaches infinity.

Problem 8 uses the squeeze theorem to evaluate a limit involving a sine function with rapidly oscillating values.

The intermediate value theorem (IVT) is applied in Problem 9 to verify continuity and find a value of c within the interval [0, 2].

Problem 9 also provides a detailed example of factoring quadratic equations to solve for the value of c.

In Problem 10, continuity of a piecewise function is discussed, and algebraic manipulation is used to find the correct value of c.

Key algebraic techniques such as factoring, substitution, and simplification of complex expressions are demonstrated throughout the quiz.

The concept of limits approaching infinity is explored in multiple problems, focusing on simplifying higher-degree terms.

Important mathematical methods, such as direct substitution and the application of the squeeze theorem, are illustrated with clear step-by-step solutions.

Transcripts

now let's start the video quiz

for each of these problems pause the

video

and work on it

once you have your answer unpause it to

see the solution so let's go ahead and

begin

for number one

if we plug in 2 notice that we will get

4 in the bottom we won't get a 0 in the

denominator of the fraction

so therefore we can use direct

substitution to get the answer

so let's replace x

with 2

so it's going to be 2 squared plus 7

times 2

plus 6

divided by 2 plus 2.

now 2 squared is 4

7 times 2 is 14

plus 6

and 2 plus 2 is 4.

now 4

14 plus 6 is 20

and 20 plus 4 that's 24

and 24 divided by

that number was supposed to be 4. 24

divided by 4 is 6.

so this is the value of the limit which

means that answer choice b

is the correct answer

number two

find the value of the limit shown below

now if we try to use direct substitution

in the denominator

we're going to get a zero

so we don't want to do that

what we need to do is factor

so how can we factor the numerator what

two numbers multiply to negative 15

but add to the middle coefficient

of positive two

this is going to be positive five and

negative three

so to factor it's gonna be x plus five

times x minus three

now on the bottom

we can factor x squared minus nine

because they're perfect squares

the square root of x squared is x and

the square root of nine

is three one of them is going to be plus

the other is going to be minus

notice that we can cancel x minus 3.

so now we can evaluate the limit as x

approaches 3

of x plus 5 divided by x plus three

so it's going to be three plus five

divided by three plus three

and so that's eight divided by six which

reduces to four divided by three if you

divide both numbers by two

so therefore d is the right answer

number three calculate the value of the

limit shown below

so whenever you have a complex fraction

what you want to do is multiply the top

and the bottom by the common denominator

the common denominator being 4x

so on the top you want to distribute 4x

times one over x

is four

four x is the same as four x over one

and you can see the x variables will

cancel

leaving behind four

so therefore we now have is the limit

as x approaches four

and on top we have positive four

now if we multiply four x by one over

four

you can see that the fours will cancel

leaving behind x

and there's a negative sign in front

so it's going to be minus x and on the

bottom we're just going to rewrite x

minus 4

and 4 minus x

now these two factors look very similar

but they're not exactly the same so what

we're going to do

is we're going to factor out a negative

1.

if we do so negative x will change into

positive x

and positive 4

will change into negative 4.

and it's at this point that we can get

rid of the x minus four

so now we have the limit

as x approaches four of negative one

divided by four x

now we can use direct substitution

so this is going to be negative 1 over 4

times 4

which gives us a final answer of

negative 1 divided by 16

which means c

is the right answer

number four

find the value of the limit

so here we have a rational function with

a square root on the top

in a situation like this

you need to multiply the top and the

bottom by the conjugate of the numerator

the conjugate is going to be the same

thing but you got to change the negative

sign into a positive sign

and whatever you do to the top you must

also do to the bottom

now on top we're going to foil the

square root of x times the square root

of x

is equal to x

the square root of x times 4 that's

going to be positive 4

square root x

and negative 4 times the square root of

x is going to be what you see here and

finally we have negative 4 times 4 which

is negative 16.

in the denominator we're not going to

foil we're just going to rewrite what we

have

so we can see that the two middle terms

add up to zero

and they're going to disappear so now

what we have is the limit as x

approaches 16

of x minus 16

divided by

x minus 16 times the square root of x

plus four

so now we can cancel x minus sixteen

and at this point

we can

replace x with sixteen so this is going

to be one divided by the square root of

sixteen plus four the square root of

sixteen is four and four plus four

is eight

so the answer is one divided by eight

which corresponds to answer choice a

number five

evaluate the limit

so we can't plug in seven

if we plug in seven it's gonna be zero

over zero which is indeterminate and we

don't know if that's equal to zero

infinity doesn't exist or one or

negative one

so we need to check the left side and

the right sided limit

so let's start with the left side as x

approaches seven from the left

let's call this f of x

so we're going to substitute a number

that's close to 7 but from the left

let's use 6.9

6.9 minus 7 is negative point one

and the absolute value of negative point

one

is positive point one and when you

divide that by negative point one

you're going to get negative one

so therefore that is the limit

as x approaches seven

from the left side it's equal to

negative one

now what about from the right side

what is the limit

as x approaches seven

from the right side

so let's plug in a number that's greater

than 7 but close to it let's try 7.1

so 7.1

minus 7

is equal to positive

0.1 and the absolute value of positive

0.1 is positive 0.1

divided by itself that's going to equal

positive 1.

so notice that the left side and the

right side do not match

so therefore the limit

does not exist

which means e is the answer

number six

what is the value of the limit of the

trigonometric function shown below

just by looking at it it's going to be 3

divided by 5

for those of you who just want a quick

answer

so it turns out answer choice b is the

right answer but

let's do some work to get our answer

so the first thing i'm going to do

is i'm going to replace

tangent

with sine divided by cosine

so tan 3x is sine 3x

divided by cosine 3x

and we still have this 5x on the bottom

so i can write it as 1 over 5x

now what i need

to have under sine 3x is a 3x i'm going

to multiply the top and the bottom by 3.

so

i'm going to trade places with the

cosine and the 3 i'm also going to move

the x in that position as well

so this is all equal to the limit as x

approaches 0

sine

three x

divided by three x

times the limit

as x approaches zero

i still have a three on top

and already move the 3x to the left so

on the bottom i have a 5

and

a cosine 3x which i'm going to write as

1 over

cosine 3x

so now i'm going to use substitution

let's say that y is equal to 3x

so therefore this limit expression

becomes the following expression the

limit as y approaches 0

of sine y

divided by y

times the limit

as x approaches zero

and here this is a function based on x

so this is gonna be one over cosine

three x

times three over five

now the limit as y approaches zero of

sine y divided by y

that's equal to one

you just need to know that formula

you can always plug in a small value of

x and you can confirm that as one

cosine zero if we replace x with zero

three times zero is still zero

cosine zero is equal to one but

i'll replace that in the next step so

what we have is one times one

times three over five

which gives us a final answer

of three divided by five

so therefore answer choice b is the

right answer

number seven

find the horizontal asymptote of the

function shown below using limits

to do so we need to find the limit

as x approaches infinity

of the function 5x plus 8x squared

divided by three x plus two x squared

plus five

now keep in mind when x becomes very

large

five x is insignificant compared to

eight x squared

if you replace x with a thousand five

thousand

is not significant to eight million

and three is not significant to two

million

so therefore this expression

becomes equivalent to the limit as x

approaches infinity of eight x squared

divided by two x squared plus five

you can only do this when x becomes very

large

mathematically it works out eight

divided by two is four

and you can cancel the x squares because

they're the same

so what we now have is four plus five

which is equal to nine

so therefore nine is the final answer

and that's going to be the horizontal

asymptote

it's an equation and y equals nine so e

is the right answer

number eight

which the following is equivalent to the

limit shown below

so go ahead and take a minute and try

this problem

now we need to know is that the sine

function oscillates between one and

negative one

this is the graph of sine x

now the only difference between sine x

and sine one over x

is how fast it oscillates

as you approach an x value of zero it

begins to oscillate faster and faster

however

the amplitude still varies between one

and negative one

so it really doesn't matter

the angle

or the fact that we have a one over x

within sine

because as x approaches zero

one over x does not exist

as x approaches zero from the left 1

over x becomes negative infinity and as

x approaches 0 from the right 1 over x

becomes positive infinity but that

doesn't matter for the sine function

because it's always going to alternate

between negative 1 and 1. so we can make

this statement

sine of one over x

will always be between

negative one and one

now if we multiply everything by x

we can get this expression negative x

is between sine

is between x

sine one over x

and it's between positive x

so x sine 1 over x is between negative x

and positive x that's what i meant to

say

now what we can use is the squeeze term

so basically the squeeze term states

that

let's call this h of x and let's say f

of x is between

h of x and g of x

if the limit as x approaches zero of

h of x and g of x if they're the same

then the limit as x approaches f of x

must also be the same

so that's the main idea behind the

squeeze term

so the limit as x approaches zero

for h of x which is really negative x

well that's equal to zero

now the limit

as x approaches zero

of g of x which in this case is positive

x

that two is equal to zero

so therefore the limit as x approaches x

sine of one over x

since it's between

negative x and x

it too

must also equal zero

so therefore b

is the right answer

according to the squeeze term

number nine

verify that the intermediate value

theorem applies to the indicated

interval

and find the value of c guaranteed by

the theorem

so

based on the intermediate value theorem

you need to know that it states that

f has to be continuous on the closed

interval

a to b

and that f of a

cannot equal f of b

and there is some number k

which

is between f of a and f b

such that

f c is equal to k

and c

has to be in the interval of a and b

so

we got to find that value of c

such that k is between f and f b

notice that we have the value of k

k is whatever number f c is equal to so

therefore k

is zero

so first we've got to show that f of a

and f of b

well we have to show that zero is

between f and f b

we have to prove that the ivt term

applies

so let's find f of a

where

a is zero

and b is two

so f of zero is going to be zero squared

plus four times zero

minus five

so f of zero is negative five

now let's calculate f of b which is f of

two

so that's gonna be two squared plus four

times two minus five

two squared is four four times two is

eight

four plus eight is twelve minus five

that's seven

so the

intermediate value theorem

does apply to the indicated interval

as we can see k which is zero

is in between negative five and seven

so now that we know that the ivt

theorem applies

we can now find the value of c so let's

set f of c equal to zero

so basically set the function f of x

replace it with zero

and find the value of x

so zero is equal to x squared plus four

x minus five

now we need to factor

two numbers that multiply to negative

five but add to positive four

are positive five and negative one

therefore we can see that x is equal to

negative five

and positive one if you reverse the

signs

if you set x plus five equal to zero x

will equal negative five and if you set

x minus one equal to zero

x will equal positive one

now one of these values

is the c value that we're looking for

so c has to be in the interval

a to b that is between zero and two

negative five is not between zero and 2

but 1 is

so therefore c

is equal to 1

which means d is the right answer

number 10

find the value of c that will make the

function continuous at x equals two

the first thing we need to do is set

these two functions

equal to each other

so seven x squared plus c x

is equal to two x cubed plus five c plus

three

now we need to find the value of c when

x is two

so let's replace x with two

and then after that

all we have to do is just algebra

two squared is four four times seven is

twenty-eight

two to the third is eight times two is

sixteen

so now let's subtract both sides by 2c

so these will cancel

on the left we have 28

and we can combine like terms 16 plus 3

is 19.

and 5c minus 2c is 3c

now let's subtract both sides by 19.

28 minus 19

is 9. so 9 is equal to 3c

so if we divide both sides by 3 we can

see that c is nine divided by three

which is three

and that is the answer

so answer choice c

is correct