If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 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?
• 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.
• 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?
• 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.
• Where can I find the "rigorous proof" of these properties?
• 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!
• At you wrote:
lim f(x)g(x) = lim f(x) * g(x)
This leads me to understand that lim ab = lim a * lim b

However, at , 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)?
• k is a constant, not a function. Hence, it doesn't approach anything - it's value is always the same.
• At , 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.
• 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.
• 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?
• it would, it would just be harder to graph. you can go to wolframalpha.com to see this.
• Concerning , 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?
• No, if one of the limit is undefined the product of the limits can't be defined
• Is there any logarithm property for limits
• 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.
• 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
• 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