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Course: Calculus, all content (2017 edition) > Unit 1
Lesson 8: Limits of combined and composite functionsLimits of combined functions
In this video, we learn how to find the limit of combined functions using algebraic properties of limits. The main ideas are that the limit of a product is the product of the limits, and that the limit of a quotient is the quotient of the limits, provided the denominator's limit isn't zero.
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Hello,
I may have a strange question but if I understand well "Limits" in maths ...
x->n means x approach "n" but never reach it. So in the last exemple how can we say the limit is "4 over 0" and say the limit does not exist ?
What am I missing ?
best regards
(Please forgive my approximate English ... I'm French)(45 votes)
There are several ways to look at this.
First, the limit law for quotient says that the lim_(x->a) F(x)/G(x) = lim_(x->a) F(x) / lim_(x->a) G(x) if lim_(x->a) G(x)≠0. In the example, lim_(x->a) G(x) = 0, therefore it doesn't exist.
Second, dividing by 0 is undefined. So it doesn't exist.
Third, x->0 means x is infinitely close to 0 but never 0. From this we can see that h(x) -> 4 (i.e. 3.9999999...) and g(x) ->0 (e.g. 0.0000001...). Now, when you divide a number by a very small number that is infinitely close to 0, you will get an infinitely large number. In other word, h(x)->4 / g(x)->0 will give us ∞. This will give us an infinite limit. An infinite limit does not exist because ∞ is not a number.(61 votes)
What is the difference between undefined, unbound, and does not exist?(18 votes)
Consider the example off(x)=0.5x+2with the interval[2,∞):
https://www.khanacademy.org/computer-programming/fx-05x-2-2/5706788113022976
The limit as x -> ∞ of f(x) is ∞, right? But technically, infinity is unquantifiable. We know its a huge number, but by definition we can never find out exactly what its value is. So it is undefined.
Now take the limit as x -> 1 of f(x). You can't because I've set the interval to be [2,∞). It simply does not exist in our function with this domain.
Now consider the interval itself:[2,∞)
Because there is a bracket ([) next to the 2, 2 is the left-most number our function can reach. f(x) is bounded in the negative direction by 2.
Because the interval after the comma is followed by a parentheses), it means f(x) will never actually reach the x coordinate of ∞. That means our function is unbounded in the positive direction.
I hope this helps!(45 votes)
I'm sure this is probably really obvious, but is there a practical reason when you would add, subtract, multiply, or divide limits? Where might I see this applied?(18 votes)
i'm late by 3 years, but still I can answer with this example:
In video games things have hitboxes and that is what is the actual thing, not the image showing up. Let's say we want the area of where to shoot the player, or rather the hitbox. For this with the help of limit relating to the hitbox/player's width + height will solve it.(11 votes)
In the third example how we can plot h(x) over g(x) graphically?(7 votes)- To my knowledge at the moment, in order to plot h(x)/g(x) you would first need the formulas for both. The example in this video does not have the formula. In other words, you cannot graph h(x)/g(x) with the information given.
EDIT: (Thanks to Eric and stolenunder)
I have just read that you actually could do it, albeit only for individual points, not for entire functions. Better explanations in the comments below.(4 votes)
At4:00, How could we create a graph of h(x)/g(x)? Or basically how we represent h(x)/g(x) and h(x)g(x)?(1 vote)
So, I've gone ahead and plotted both functions with some linear equations (Just the second example):
h(x): https://www.desmos.com/calculator/pu0ceau0hh (The line from x = 0 to x = 3 isn't straight here. I had to make it bend as otherwise, my intervals wouldn't match between functions and I wouldn't be able to divide the functions)
g(x): https://www.desmos.com/calculator/fqenm4puix (This one is pretty much exact)
h(x)/g(x): https://www.desmos.com/calculator/oute3jh0ys
As you can see, the limit as x goes to 0 is infinity here, which is in line with the fact that in the video, Sal mentioned the limit does not exist (A limit will not exist when the function either approaches infinity, -infinity, or it approaches two different values from both sides). So, I think that's proven.
Anyway, this isn't something you should be doing in a test (for very obvious reasons😂). This was just to give you an idea.
Also, I removed my previous answer, considering I was able to model the function after all(13 votes)
Sal, you use desmos right?(5 votes)
Yes, Khan Academy uses Desmos to graph most of their functions.(1 vote)
- Hello guys! Shouldn't the final result be = ∞?
I mean, if the numerator approaches a constant and the denominator approaches zero, the fraction should approach infinity, right?
For instance:
http://www.wolframalpha.com/input/?i=lim+(4%2Bx)%2F%7Cx%2F4%7C+x-%3E0
(Copy the full URL in order to open the link properly.)
Any help is welcomed :)(4 votes)
999 over 1 is a big number, 999 over 0 is undefined.
You are right, the graph the link shows does approach infinity. However, infinity is not a real number. So in this case, the limit is undefined.(2 votes)
- At1:56, the limit of f(x) does not exist as x approaches to 0 since it is open at x=0 i.e., f(0) does not exist still you have considered the limit exist and taken its value as -1. Why?(3 votes)
- This is because, it is apparent that as x approaches 0, the function would output -1 IF it were continuous. This is an assumption that has to be made when calculating limits of any function. Though remember that when calculating the limit at a jump discontinuity you can only find the limit from above or below. Cheers!(4 votes)
- It was my understanding that the limits of functions at sharp or drastically changing points do not exist? is that right or wrong?(2 votes)
What do you mean by sharp or drastic? Something like |x| at x=0?, or |cos x| at, for example π/2?
https://www.desmos.com/calculator/311wvcopnj
At those points, the limit does exist, that is, the left and right limits are equal. However the function is not differentiable there. So in these cases, you are wrong.
But if you are meaning something more like a piecewise function, say f(x) = {x=-1 for x -1 < x < 1 else x=1}, does the limit of x=-1 or x=1 exist? In this case, yes, you are correct, the limits at x=-1 and x=1 do not exist.
https://www.desmos.com/calculator/dswoobhzzg(5 votes)
whats the difference between combined and composite functions?(2 votes)
Combined functions are functions merged together using arithematic +×÷-
Composite is where you insert the reflection of 1 function into the other.
f(g(x))=?
How to look at this:
What is g(x)? g(x) is the y value when you insert an x into the equation [g(x) = (blah)]
You then take that y value and insert it as the x value for the function f(x)(5 votes)
4:00Video transcript
- [Instructor] Let's
find the limit of f of x times h of x as x approaches zero. All right, we have graphical depictions of the graphs y equals f
of x and y equals h of x. And we know, from our limit properties, that this is going to be
the same thing as the limit as x approaches zero of f of x times, times the limit as x approaches zero of h of x. And let's think about
what each of these are. So let's first think about
f of x right over here. So on f of x, as x approaches zero, notice the function itself
isn't defined there. But we see when we approach from the left, we are approaching the, the function seems to
be approaching the value of negative one right over here. And as we approach from the
right, the function seems to be approaching the
value of negative one. So the limit here, this limit here is negative one. As we approach from the left, we're approaching negative one. As we approach from the right,
the value of the function seems to be approaching negative one. Now what about h of x? Well, h of x we have down here. As x approaches zero, as x approaches zero, the function is defined at x equals zero. It looks like it is equal to one. And the limit is also equal to one. We can see that, as we
approach it from the left, we are approaching one. As we approach from the
right, we're approaching one. As we approach x equals zero
from the left, we approach, the function approaches one. As we approach x equals
zero from the right, the function itself is approaching one. And it makes sense that the
function is defined there, is defined at x equals zero, and the limit as x approaches zero is equal to the same as the, is equal to the value of
the function at that point because this is a continuous function. So this is, this is one. And so negative one times one is going to be equal to, is equal to negative one. So that is equal to negative one. Let's do one more. All right, so these are both, looks like continuous functions. So we have the limit as x
approaches zero of h of x over g of x. So once again, using our limit properties, this is going to be the same
thing as the limit of h of x as x approaches zero over the limit of g of x as x approaches zero. Now what's the limit of h
of x as x approaches zero? This is, let's see, as we approach zero from the left, as we approach x equals
zero from the left, our function seems to be approaching four. And as we approach x
equals zero from the right, our function seems to be approaching four. That's also what the value of the function is at x equals zero. That makes sense because this
is a continuous function. So the limit as we approach x
equals zero should be the same as the value of the
function at x equals zero. So this top, this is going to be four. Now let's think about the
limit of g of x as x equals, as x approaches zero. So from the left, it looks
like, as x approaches zero, the value of the function
is approaching zero. And as x approaches zero from the right, the value of the function
is also approaching zero, which happens to also be, which also happens to be g of zero. G of zero is also zero. And that makes sense that the
limit and the actual value of the function at that point is the same because it's continuous. So this also is zero, but now we're in a strange situation. We have to take four
and divide it by zero. So this limit will not exist 'cause we can't take four
and divide it by zero. So even though the limit
of h of x is x equals, as x approaches zero exists and the limit of g of x as
x approaches zero exists, we can't divide four by zero, so this whole entire limit does not exist, does not exist. And actually, if you were
to plot h of x over g of x, if you were to plot that graph, you would see it even clearer that that limit does not exist. You would actually be able
to see it graphically.