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### Course: Class 12 math (India)>Unit 5

Lesson 2: Review: Limits of combined & composite functions

# Limits 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.

## Want to join the conversation?

• 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)
• 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.
• What is the difference between undefined, unbound, and does not exist?
• Consider the example of `f(x)=0.5x+2` with the interval`[2,∞)`:

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!
• 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?
• 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.
• In the third example how we can plot h(x) over g(x) graphically?
• 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.
• At , 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
• Sal, you use desmos right?
• 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 :)
• 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.
• At , 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?
• 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!
• It was my understanding that the limits of functions at sharp or drastically changing points do not exist? is that right or wrong?
• 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
• whats the difference between combined and composite functions?