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## Calculus, all content (2017 edition)

### Unit 1: Lesson 8

Limits of combined and composite functions# Limit properties

AP.CALC:

LIM‑1 (EU)

, LIM‑1.D (LO)

, LIM‑1.D.1 (EK)

, LIM‑1.D.2 (EK)

What is the limit of the sum of two functions? What about the product? Created by Sal Khan.

## Want to join the conversation?

- I seen in my book the notation like lim f(x+y). Is this the same as f(x)+f(y) ?

Also my teacher said that ln(x+y) is not lnx + lny but ln(x,y) = ln(x)+ln(y). Does ln represent a function?(2 votes)- Yes, "ln" represents a function. Ln(x) means the natural logarithm of x. The natural logarithm of x is what power you have to raise the constant e (2.7831...) to equal x.(4 votes)

- What I do not understand about limits is why you would want to do something like multiply or add them? How can limits be related to each other, let alone multiplied? Aren't limits simply in respect to themselves as a limit, and not in respect to other limits?(53 votes)
- These properties of limits give you tools for finding the limit of a more complicated function. You can break down a function into different parts (as defined by the limit properties) and find the limits of the different parts, putting them back together in the appropriate manner to find the limit of the overall function. For example the limit of:

5x^3 + 4x^2 + 3x - 6

can be found by breaking it up into 5 times (limit of x)^3 plus 4 times (limit of x)^2 plus 3 times (limit of x) - 6. So you only have to find the limit of one thing, x, and then apply the appropriate limit properties. This will become even more powerful when Sal introduces how to find the limit of a composite function.(2 votes)

- Where can I find the "rigorous proof" of these properties?(10 votes)
- A rigorous proof can usually be found in any old calculus text, in the section on limits. A fun exercise might be to write down the epsilon-delta definition of limits then try to figure out exactly how one would prove these statements!(7 votes)

- At2:21you wrote:

lim f(x)g(x) = lim f(x) * g(x)

This leads me to understand that lim ab = lim a * lim b

However, at3:22, you wrote:

lim kf(x) = k * lim f(x)

If lim ab = lim a * lim b, shouldn't lim kf(x) = lim k * lim f(x)?(4 votes)- k is a constant, not a function. Hence, it doesn't approach anything - it's value is always the same.(15 votes)

- At3:52, note that the lim of g(x) as it approaches c can not be 0. If it is, then the entire limit would not exist because a number divided by 0 is undefined.(10 votes)
- My real analysis textbook says that (and I will paraphrase) if the limits of f(x) and g(x) exist as x approaches c and the limit of g(x) as x approaches c is not equal to 0, THEN the limit of (f/g)(x)=(lim f(x))/(lim g(x)) as each of the limits approaches 0.

I think this is what grant was getting at, but the video didn't go into detail about this.

Also, Just Keith, your example with the limit is one where x approaches 0 and the answer is infinity, but if that was our g(x), it would not be the case that the limit of g(x) as x approaches c is 0. That said, it really does not apply to this situation. I see what your point is, but I think some of your response was fairly unrelated to the original comment.(2 votes)

- for the exponent property, why assume that the power is a rational number? would this still work if the power was sqrt(3) or something like that?(6 votes)
- it would, it would just be harder to graph. you can go to wolframalpha.com to see this.(7 votes)

- Concerning2:20, If you get the question "what is the product of the limit g(x) when x approaches c and the limit f(x) when x approaches c" given that f(x) is discontinuous at c. Would the answer be zero or just that the limit does not exist?(4 votes)
- No, if one of the limit is undefined the product of the limits can't be defined(8 votes)

- Is there any logarithm property for limits(4 votes)
- The closest thing to a 'logarithm property' is the rule regarding continuous functions. The limit of f(g(x)) is equal to f(the limit of g(x)), provided f is continuous at that limit. Logarithms are continuous on their domain, so we can apply that to say lim (ln(f(x))) = ln (lim f(x)) for a positive inner limit. We can also say ln(lim f(x)) = lim ln(f(x)), which is occasionally useful.(3 votes)

- I was given this problem in one of my textbooks about limits but do not understand the answer given. Find lim as h--->0 of (e^2(3+h)-e^2(3))/h by recognizing the limit as the definition of f (a′) for some function f and some value a.

the answer given is f'(3) where f(x)=e^2x. f'(3)=2e^6 would appreciate any input on how they came to this answer

thanks(2 votes)- You don't plug in the value of x until after you have done the limit. So, you should do the limit calculation with the x instead of the 3. Once you have the limit in terms of x, then you plug in the x=3.

Here is how to do the limit, if I understand the example correctly.

f(x) = e^(2x)

f'(x)= lim h→0 {e^[2(x+h)] - e^(2x) } / h

f'(x) = lim h→0 {e^[2x+2h] - e^(2x) } / h

f'(x) = lim h→0 {e^(2x)∙e^(2h)] - e^(2x) } / h

Factor out e^(2x). Since it is not a function of h, we can factor it in front of the limit

f'(x) = lim h→0 e^(2x) {e^(2h)] - 1 } / h

f'(x) = e^(2x) lim h→0 {e^(2h) - 1 } / h

-----

Side calculation:

Definition of e is lim h→0 (1+h)^(1/h)

Thus, e^(2h) = lim h→0 [ (1+h)^(1/h)]^(2h)

e^(2h)= lim h→0 (1+h)^(2h/h)

e^(2h)= lim h→0 (1+h)^(2)

e^(2h)= lim h→0 (1 + 2h + h²)

Since we both limits have the limit variable approaching 0, we can substitute the e^(2h) in main calculation with (1 + 2h + h²)

------

Back to main calculation:

f'(x) = e^(2x) lim h→0 {e^(2h) - 1 } / h

f'(x) = e^(2x) lim h→0 { 1 + 2h + h² - 1 } / h

f'(x) = e^(2x) lim h→0 {2h + h² } / h

f'(x) = e^(2x) lim h→0 h(2 + h) / h

cancel out the h

f'(x) = e^(2x) lim h→0 (2 + h)

Apply the limit:

f'(x) = e^(2x) (2 + 0)

f'(x) = e^(2x) (2)

f'(x) = 2e^(2x)

Now, apply x=3

f'(3) = 2e^(2∙3) = 2e^6(7 votes)

- What is the limit of f(x) raised to the g(x) power?(4 votes)
- Assuming that the limits of f(x) and g(x) are defined and non-zero for f(x), then you can just substitute.

If direct substitution leads to an indeterminate form§, the short answer is that to figure this out you convert the power into an exponential function and then bring the limit inside the exponential function.

(The latter can be done because the exponential function is continuous.)

This converts that into a form that allows L'Hôpital's rule to be applied.

A more complete explanation and examples can be found here:

https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/power_limits.html

§Note: For a description of indeterminate forms see:

https://en.wikipedia.org/wiki/Indeterminate_form(2 votes)

## Video transcript

What I want to do in
this video is give you a bunch of properties of limits. And we're not going to
prove it rigorously here. In order to have the rigorous
proof of these properties, we need a rigorous definition
of what a limit is. And we're not doing
that in this tutorial, we'll do that in the
tutorial on the epsilon delta definition of limits. But most of these should
be fairly intuitive. And they are very helpful for
simplifying limit problems in the future. So let's say we know that
the limit of some function f of x, as x approaches
c, is equal to capital L. And let's say that we
also know that the limit of some other function, let's
say g of x, as x approaches c, is equal to capital M. Now given that, what
would be the limit of f of x plus g of x
as x approaches c? Well-- and you could
look at this visually, if you look at the graphs
of two arbitrary functions, you would essentially just
add those two functions-- it'll be pretty clear that
this is going to be equal to-- and once again, I'm not
doing a rigorous proof, I'm just really giving
you the properties here-- this is going to be the limit
of f of x as x approaches c, plus the limit of g of
x as x approaches c. Which is equal to, well
this right over here is-- let me do that
in that same color-- this right here is
just equal to L. It's going to be equal to L
plus M. This right over here is equal to M. Not too difficult. This is often called the sum
rule, or the sum property, of limits. And we could come up with a very
similar one with differences. The limit as x approaches
c of f of x minus g of x, is just going to be
L minus M. It's just the limit of f of
x as x approaches c, minus the limit of g
of x as x approaches c. So it's just going
to be L minus M. And we also often
call it the difference rule, or the difference
property, of limits. And these once again, are very,
very, hopefully, reasonably intuitive. Now what happens if you take
the product of the functions? The limit of f of x times
g of x as x approaches c. Well lucky for us,
this is going to be equal to the limit of
f of x as x approaches c, times the limit of g
of x, as x approaches c. Lucky for us, this is kind of
a fairly intuitive property of limits. So in this case,
this is just going to be equal to, this is
L times M. This is just going to be L times
M. Same thing, if instead of having a function
here, we had a constant. So if we just had
the limit-- let me do it in that same
color-- the limit of k times f of x, as x approaches c,
where k is just some constant. This is going to be the same
thing as k times the limit of f of x as x approaches c. And that is just equal
to L. So this whole thing simplifies to k times L. And we can do the same
thing with difference. This is often called the
constant multiple property. We can do the same
thing with differences. So if we have the
limit as x approaches c of f of x divided by g of x. This is the exact same
thing as the limit of f of x as x
approaches c, divided by the limit of g of
x as x approaches c. Which is going to be equal
to-- I think you get it now-- this is going to be
equal to L over M. And finally-- this is sometimes
called the quotient property-- finally we'll look at
the exponent property. So if I have the
limit of-- let me write it this way-- of
f of x to some power. And actually, let
me even write it as a fractional
power, to the r over s power, where both r and s are
integers, then the limit of f of x to the r over s
power as x approaches c, is going to be the exact
same thing as the limit of f of x as x approaches c
raised to the r over s power. Once again, when r and s
are both integers, and s is not equal to 0. Otherwise this exponent
would not make much sense. And this is the same thing
as L to the r over s power. So this is equal to L
to the r over s power. So using these, we
can actually find the limit of many,
many, many things. And what's neat about it is
the property of limits kind of are the things that you
would naturally want to do. And if you graph some
of these functions, it actually turns out
to be quite intuitive.