Calculus 1 - Introduction to Limits
Summary
TLDRThis 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
📘 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.
🔢 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.
📉 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.
📊 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.
🚫 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.
Mindmap
Keywords
💡Limits
💡Direct Substitution
💡Indeterminate Form
💡Factoring
💡Conjugate
💡Vertical Asymptote
💡Jump Discontinuity
💡Hole
💡Graphical Evaluation
💡One-Sided Limits
💡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
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