Limits and Continuity
Summary
TLDRThis 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
🧑🏫 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.
🧮 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.
📐 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.
🧮 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.
Mindmap
Keywords
💡Limit
💡Direct Substitution
💡Factoring
💡Indeterminate Form
💡Conjugate
💡Intermediate Value Theorem (IVT)
💡Squeeze Theorem
💡Horizontal Asymptote
💡Continuous Function
💡Trigonometric Limits
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
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