Calculus 1 - Introduction to Limits

Summary

This video offers a comprehensive introduction to evaluating limits in calculus, both analytically and graphically. It covers direct substitution, factoring, and using conjugates for complex fractions and radicals. The script guides viewers through various examples, illustrating how to find limits as x approaches different values and explains one-sided limits, function values, and different types of discontinuities such as vertical asymptotes, jump discontinuities, and holes.

Takeaways

• 📘 Direct substitution is a method to find limits by plugging in values close to the point of interest, but not the point itself.
• 🔍 When direct substitution results in an undefined expression (like 0/0), other techniques such as factoring are necessary to simplify the function.
• 🔢 Factoring expressions, especially differences of squares and cubes, can help simplify limits by eliminating problematic terms like zero in the denominator.
• 📉 For limits involving fractions, cancelling common factors in the numerator and denominator can simplify the expression and make direct substitution possible.
• 📈 The limit of a function can be evaluated graphically by observing the behavior of the graph as it approaches a certain point from the left or right.
• 📊 Left-sided and right-sided limits can differ, indicating a discontinuity at a point, and if they match, the limit exists from either side.
• 🚫 Vertical asymptotes occur when the function has a zero in the denominator at a certain point, making the function undefined there.
• 🔄 Complex fractions and radicals can be handled by multiplying by the common denominator and the conjugate to simplify the expression.
• 📋 The value of the function at a certain point is found by looking for the y-value at the corresponding x-coordinate on the graph, which may differ from the limit.
• ⚠️ Discontinuities such as jump discontinuities and holes can affect the existence of limits and the function's value at specific points.

Q & A

What is the main topic of the video?

The main topic of the video is an introduction to limits, focusing on how to evaluate them both analytically and graphically.

What is the first example given in the video to explain limits?

The first example is finding the limit as x approaches two of the function (x^2 - 4) / (x - 2).

Why is direct substitution not possible for the first example in the video?

Direct substitution is not possible because plugging in x = 2 results in a 0/0 indeterminate form, which is undefined.

How does the video suggest finding the limit when direct substitution fails?

The video suggests plugging in values close to the point of interest but not exactly that point, or using algebraic manipulation such as factoring.

What is the limit of the function (x^2 - 4) / (x - 2) as x approaches 2 according to the video?

The limit is 4, as demonstrated by plugging in values close to 2 and observing the function's behavior.

What technique does the video use to simplify the first example before finding the limit?

The video uses factoring to simplify the expression (x^2 - 4) as (x + 2)(x - 2) and then cancels out the (x - 2) term before substitution.

How does the video handle limits involving fractions where the denominator becomes zero?

The video suggests factoring the numerator and denominator to cancel out the common factors causing the zero in the denominator.

What is the limit of the function x^3 - 27 / (x - 3) as x approaches 3, and what technique is used?

The limit is 27, and the technique used is factoring the numerator as a difference of cubes and then canceling the common (x - 3) factor.

How does the video approach evaluating limits graphically?

The video demonstrates evaluating limits graphically by looking at the y-values as x approaches a certain point from the left or right side of a graph.

What are the different types of discontinuities mentioned in the video?

The video mentions jump discontinuities, infinite discontinuities, and holes (removable discontinuities).

How does the video determine if a limit exists graphically?

The video determines if a limit exists graphically by checking if the left-sided limit and the right-sided limit match at a given point.

Outlines

00:00

📘 Introduction to Limits

This paragraph introduces the concept of limits in calculus, explaining how to evaluate them both analytically and graphically. The example provided is the limit of the function (x^2 - 4) / (x - 2) as x approaches 2. The concept of direct substitution is introduced, and it's shown that plugging in x=2 leads to an undefined result due to division by zero. The technique of plugging in values close to 2, such as 1.9 and 2.1, is demonstrated to approximate the limit. Factoring the numerator to cancel out the (x - 2) term in the denominator is also discussed, which simplifies the expression and allows for direct substitution to find the limit as x approaches 2, which is 4.

05:01

🔢 Analyzing Limits with Direct Substitution and Factoring

This section explores direct substitution in limits where there is no division by zero, exemplified by finding the limit of x^2 + 2x - 4 as x approaches 5. The process of factoring expressions to simplify limits, particularly when dealing with fractions that could lead to division by zero, is explained. The example of the limit of (x^3 - 27) / (x - 3) as x approaches 3 is used to illustrate the factoring of a difference of cubes and how it leads to a determinable limit of 27.

10:01

📉 Evaluating Limits with Complex Fractions and Radicals

The paragraph discusses strategies for evaluating limits involving complex fractions and radicals. It suggests multiplying the numerator and denominator by a common denominator and the conjugate of the radical expression to simplify the expression. Examples include simplifying 1/(x - 1) / (x - 3) by multiplying by 3x and the conjugate, leading to a determinable limit. Another example involves the limit of √(x - 3) / (x - 9), where multiplying by the conjugate simplifies the expression and allows for the limit to be evaluated as x approaches 9.

15:04

📊 Graphical Interpretation of Limits

This section explains how to evaluate limits graphically by examining the behavior of a function as it approaches a certain point. It describes how to find the left-sided and right-sided limits and how they can differ, leading to the conclusion that the limit does not exist if they are not equal. Examples include approaching x = -3 from the left and right, illustrating how the function's behavior can lead to different y-values, indicating the non-existence of the limit at that point. The concept of function values at specific points, vertical asymptotes, jump discontinuities, and holes in the graph are also introduced.

20:05

🚫 Discontinuities and Vertical Asymptotes

The final paragraph delves into the concept of discontinuities in functions, specifically focusing on jump discontinuities and infinite discontinuities, which are non-removable. It contrasts these with removable discontinuities, such as holes in the graph. The discussion includes examples of functions with vertical asymptotes, where the function value is undefined at certain points, and how these are represented graphically.

Keywords

💡Limits

Limits in calculus refer to the value that a function or sequence 'approaches' as the input or index approaches some value. In the video, limits are the central theme, with the focus on evaluating them both analytically and graphically. For instance, the script discusses finding the limit of a function as x approaches a certain value, like in the example where the limit of (x^2 - 4) / (x - 2) as x approaches 2 is evaluated.

💡Direct Substitution

Direct substitution is a method used to evaluate limits by simply substituting the value of x into the function. However, as mentioned in the script, if substitution results in an indeterminate form like 0/0, it means the limit cannot be found this way. The video uses direct substitution in examples where it's applicable, such as evaluating the limit of x^2 + 2x - 4 as x approaches 5.

💡Indeterminate Form

An indeterminate form arises in limits when direct substitution results in an expression that does not provide a specific value, like 0/0 or ∞/∞. The script explains that when this happens, as with the function (x^2 - 4) / (x - 2) when x approaches 2, other techniques must be used to find the limit.

💡Factoring

Factoring is a technique used to simplify expressions, especially those that result in indeterminate forms when evaluating limits. The video demonstrates how factoring x^2 - 4 into (x + 2)(x - 2) allows for the cancellation of terms and the subsequent use of direct substitution to find the limit as x approaches 2.

💡Conjugate

The conjugate of an expression is used to eliminate square roots or other radicals when evaluating limits. The script illustrates multiplying by the conjugate in examples involving square roots, such as multiplying the numerator and denominator by the conjugate to simplify the expression and find the limit as x approaches a certain value.

💡Vertical Asymptote

A vertical asymptote is a vertical line on a graph where the function approaches infinity or negative infinity. The video explains that a vertical asymptote occurs where the function has a zero in the denominator and cannot be evaluated, such as at x = 3 in the function 1/(x - 3).

💡Jump Discontinuity

A jump discontinuity is a type of discontinuity in a function where there is a sudden 'jump' in the value of the function. The video describes a jump discontinuity at x = -3, where the graph does not connect, indicating the function has different left and right limits that do not match.

💡Hole

A hole in a function's graph is a point where the function is not defined, but the limits from both sides exist and are equal. The video mentions a hole at x = -2, where the function has a removable discontinuity, meaning the hole can be 'filled in' by redefining the function at that point.

💡Graphical Evaluation

Graphical evaluation of limits involves looking at the graph of a function to determine the value the function approaches as x approaches a certain value. The video provides examples of evaluating limits from the left and right sides graphically, and how to determine if a limit exists based on whether these values match.

💡One-Sided Limits

One-sided limits are limits evaluated as the input approaches a certain value from either the left or the right. The video explains how to find these by looking at the graph from the appropriate direction and gives examples where one-sided limits exist or do not exist.

💡Removable Discontinuity

A removable discontinuity is a type of discontinuity that can be 'removed' by redefining the function at the point of discontinuity. The video contrasts this with a jump discontinuity, showing that a hole at x = -2 is an example of a removable discontinuity.

Highlights

Introduction to limits and their evaluation methods.

Direct substitution method for evaluating limits.

Undefined limits and the importance of approaching the value without substitution.

Technique of plugging in values close to the limit point to find the limit.

Factoring as a method to simplify limits involving fractions.

Cancellation of terms to simplify the limit expression.

Direct substitution after factoring to find the limit of a function.

Evaluating limits of polynomial functions without fractions.

Handling limits involving difference of cubes using algebraic identities.

Graphical evaluation of limits by approaching a point from the left and right.

Understanding one-sided limits and their significance.

Identifying vertical asymptotes and their impact on limit evaluation.

Dealing with complex fractions by multiplying by the common denominator.

Simplifying expressions involving square roots by multiplying by the conjugate.

Graphical identification of jump discontinuities and their effect on function values.

Differentiating between removable and non-removable discontinuities.

Practical examples of evaluating limits graphically and analytically.

Transcripts

in this video we're just going to go

over a basic introduction into limits

and how to evaluate them analytically

and graphically

so here's a simple example

let's say if we want to find the limit

as x approaches two

of the function x squared minus four

divided by x minus two

so how can we do so

well one way is to use direct

substitution

if we plug in two

notice what will happen

two squared is four four minus four is

zero

so zero over zero is undefined

which

we don't know what value that represents

now sometimes

you could find the limit by plugging a

value that's close to two

and that's what you want to do you want

to plug in a number that's close to two

but not exactly two

so for example

let's call this f of x

so let's calculate f of 1.9

and let's see

what's going to happen

actually let's make it 2.1

so let's get a positive answer instead

of a negative one

two point one squared

minus 4

that's about

0.41

and 2.1 minus 2 is 0.1

so this is going to be

4.1

now what if we pick a value that's even

closer to 2 for example

let's try 2.01

so if you type this in the way you see

it in your calculator you may have to

put this in parenthesis

you should get

4.01

so notice what's happening

as we get closer and closer to two

the limit approaches four

so we could therefore say that the limit

as x approaches two of this function

is equal to four

and this technique works for any limit

as long as you plug in a number that's

very close to whatever this number is

but not exactly that number

if the limit exists it's going to

converge to a certain value

now sometimes

you have to use other techniques to get

the answer

in this particular example

we could factor

x squared minus four you can write it as

x plus two

times x minus two

now we need to rewrite the limit

expression

until we replace x with two

now notice that we can cancel x minus

two

so when this term is gone

we can now use direct substitution

because

the x minus two factor was giving us a

zero in the denominator which we don't

want so now all we need to do is find

the limit

as x approaches two of x plus two

so now we can replace x with two and two

plus two is four

and so that's the limit

it approaches a value of 4.

now let's look at another example

what is the limit

as x approaches 5

of x squared plus two x minus four

so notice that we don't have a fraction

we're not going to get a zero in the

denominator so for a question like this

you can use direct substitution so all

you got to do is plug in five

so it's going to be five squared plus

two times five

minus four

so that's 25 plus 10

minus four

and 25 plus 10 is 35.

so the limit is going to be 31.

and so that's it for that example

but now what about this one what is the

limit

as x approaches 3

of x cubed minus 27

over x minus 3.

now if we try to plug in 3 it's going to

be 0 over 0 so we don't want to do that

in this case if you have a fraction like

this see if you can factor

the expression

so how can we factor x cubed minus 27

so what we have is a difference of cubes

and whenever you see that you can use

this formula aq minus b cubed

is a minus b

times a squared plus

a b

plus b squared

so in our example

a to the third is like x to the third

and b to the third is 27.

so a is the cube root of x cubed which

is x

and b is the cube root of 27 which is

three

so this is going to be x minus 3

and then a squared that's x squared

and then plus a b so that's 3 times x

and then plus b squared or 3 squared

which is 9.

so now we can cancel the factor x minus

three

so what we have left over is the limit

as x approaches three

of x squared plus three x plus nine

so at this point we now can use direct

substitution

so it's three squared plus

three times three

plus nine

which is nine plus nine plus nine

adding nine three times is basically

multiplying nine by three

and so this limit is equal to

twenty-seven

now here's another problem that you can

work on

so what is the limit

as x approaches

three

of one over x minus one over three

divided by x minus three

so for these examples feel free to pause

the video if you want to and try these

problems

so in this example we have a complex

fraction

so what do you do in a situation like

this

if you get a complex fraction what i

recommend is to multiply the top and the

bottom

by the common denominator of those two

fractions

that is by x and by 3. so i'm going to

multiply the top and the bottom

by 3x

so if we multiply 3x by 1 over x

the x variables will cancel

and so what we're going to have left

over

is simply 3

and if we multiply 3x by 1 over 3

the 3s will cancel leaving behind x but

there's a negative sign in front of it

now for the terms on the bottom i'm

going to leave it in its factored form

so notice that 3 minus x and x minus 3

are very similar

if you see a situation like this

factor out a negative one

if we take out a negative one the

negative x

will change to positive x

and positive three

will change to negative three

so notice that we can cancel

the x minus 3 factor at this point

and so what we have left over

is the limit as x approaches 3

of negative 1 over three x

so now we can use direct substitution

so let's replace x with three

so it's going to be three times three

which is nine so the final answer

is negative one

divided by nine

so now you know how to evaluate limits

that are associated with complex

fractions

so here's another example

what is the limit

as x approaches

nine

of square root x minus three

over x minus nine

so what should we do if we're dealing

with square roots

what i recommend

is to multiply the top and the bottom by

the conjugate

of the expression that has the square

root

so the conjugate of square root x minus

3 is square root x plus 3.

so in the numerator we need to foil

so the square root of x times the square

root of x

is the square root of x squared which is

simply x

and then we have the square root of x

times three

so that's going to be plus

three square root x and then these two

will form

negative three square root x

and finally we have negative three times

positive three which is negative nine

now on the bottom

i'm not going to foil the two

expressions i'm gonna leave it the way

it is because my goal is to get rid of

the x minus nine i want to cancel it

so now negative three and positive three

add up to zero

so what we have left over is the limit

as x approaches 9

of x minus 9

divided by

x minus 9 times the square root of x

plus 3.

so now at this point notice that we can

cancel x minus nine

and so what we have left over

is the limit as x approaches nine

of one over

square root x plus three

so now what we can do

is replace x with nine so i'm just going

to continue up here

so it's going to be 1 over square root 9

plus 3

and the square root of 9

is 3

and 3 plus three is six so the final

answer

is one over six

now let's look at this example

what is the limit as x approaches four

of the expression

one over square root x

minus one over two

divided by x minus four

so this time we have a complex fraction

with

radicals

that means we need to multiply the top

and the bottom not only by the common

denominator but also by the conjugate

but let's start with a common

denominator so i'm going to multiply the

top and the bottom

by these two that is

by 2 square root x

so when i multiply 1 over square root x

times 2 square root x

the square root x terms will cancel

leaving behind positive two so i have

the limit

as x approaches four

with a two on top

and if i multiply these two the twos

will cancel leaving behind the square

root of x

and so on the bottom i have 2 square

root x

times x minus 4.

so now my next step

is to multiply the top and the bottom by

the conjugate

of the radical expression that is by 2

plus

square root x

now i'm only going to foil the top part

not the bottom

so we have 2 times 2

which is 4

and then we have 2 times the square root

of x

and then these two that's going to be

negative 2 square root x

and negative square root x times

positive square root x

is going to be

negative square root x squared which is

negative x

and in the denominator

don't foil just rewrite what you have

if you follow these steps

it won't be that difficult so now 2 and

negative 2 adds up to 0.

so what we have left over is the limit

as x approaches 4

of 4 minus x

divided by

all the stuff that we have on the bottom

now what we're going to do at this point

is we're going to factor out a negative

one

so this is now the limit

as x approaches four

so negative one and

this is going to change to positive x

and plus 4 is going to change to

negative 4 just like we did before

so now notice that we can cancel

the x plus four terms at this point

so this is what we have left over

so now let's replace x with four

so it's negative one

divided by two

square root four

times two plus

square root four

so the square root of four

is two

and two times two

that's going to be four and two plus two

is four

and four times four is sixteen

so the final answer is negative one

divided by sixteen

and so that's going to be the limit

now let's talk about how to evaluate

limits graphically

so let's say

if we want to calculate the limit

as x approaches

negative 3

from the left side

and let's say this graph represents the

function f of x

so what can we do

so to evaluate the limit you're looking

for the y value so first identify where

x is negative three

x is negative 3 anywhere along that

vertical line

now we want to approach that vertical

line

from the left side

so therefore you want to follow the

curve

from the left until you get to that

point

so notice that the y value here

corresponds to positive one

so therefore the limit

as x approaches three from the left side

is one

now what about from the right side

so you want to approach the

vertical line at negative 3 from the

right

so we got to follow this curve

so notice that the y value here

is negative 3.

so therefore the

limit as x approaches negative three

from the right side

that's a negative three

now what about the limit

as x approaches negative three from

either side

if the left-sided limit and the

right-sided limit are not the same then

the limit does not exist

so these two are known as one-sided

limits

now what about f of negative three

what is the value of the function when x

is negative three

to find it identify the closed circle

which has a y value of negative three so

this is it

now let's work on some more problems

so what is the limit

as x approaches

negative two from the left side

go ahead and try this one

so identify the vertical line at

negative two so we want to approach that

line

from the left side

so notice that the y value

is negative two

so what is the limit

as x approaches negative two from the

right side

so this time

we want to approach the vertical line

from the right

and it points to the same value

negative two

so therefore the limit

as x approaches negative two from either

side

does exist

because these two they match

so therefore it's going to be negative

two now what is the function value when

x is negative 2

so look for the closed circle

that's on this vertical line

and so that's this point where the y

value is positive 2.

so as you can see

is not very difficult to evaluate limits

graphically

now let me give you a new set of

problems

evaluate the limit

as x approaches positive one

from the left side

from

the right side

and

from

either side

and also find the function value when x

is one

so

x is one anywhere along that vertical

line so if we approach it from the left

side notice that the y value

is positive one there

and if we approach it from the right

side

the y value is three

so because these two do not match

the limit as x approaches one

does not exist

now the function value at one is the

closed circle

which has a y value of two

now here's the last set

of problems like this

what is the limit

as x approaches positive 3

from the left side

and

from the right side

and from either side

and then find the function value at 3.

so notice that at x equals 3 we have

this vertical asymptote

so as x approaches 3 from the right i

mean from the left side

notice that it goes down to negative

infinity

and as we approach 3 from the right side

it goes all the way up

to positive infinity

now these two do not match

so therefore the limit does not exist

and the function value at 3 is going to

be undefined

so a good example of having a vertical

asymptote at x equals 3 would be a

function like this one over x minus

three

and if you plug in three you're gonna

get one over zero

which is undefined

so whenever you have a zero in the

denominator

at that point you have a vertical

asymptote

and it's undefined at that point

now at negative 3

we have what is known as

a jump discontinuity the graph doesn't

connect

at negative two we have what is known as

a hole

a hole is a removable discontinuity

a jump discontinuity is not removable

so this is another example of a jump

discontinuity it's a non-removable

discontinuity

and here we have an infinite

discontinuity which is also

non-removable