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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 1

Lesson 5: Determining limits using algebraic properties of limits: limit properties

# Limits of combined functions

Sal solves a few examples where the graphs of two functions are given and we're asked to find the limit of an expression that combines the two functions. For example, the limit at 0 of the product of the functions.

## 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? • 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 , 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? • 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 :) • It was my understanding that the limits of functions at sharp or drastically changing points do not exist? is that right or wrong? • When applying limit rules for f(x) and h(x) for example, do the functions have to both be continuous? • whats the difference between combined and composite functions? • 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)
• Define f(0), so that the function, f(x)=((1-x)^n )-1/ x , n epsilon N, is continuous at X=0
I'm stuck on this question, because I'm still a little confused. the concept isn't very clear to me yet 