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## Precalculus

### Course: Precalculus>Unit 10

Lesson 4: Determining limits using algebraic properties of limits: limit properties

# Limit properties

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? • 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? •   They can be. For instance, let's say that I had a rectangle made out of metal that expanded or contracted depending on the temperature. If I knew that the length of the rectangle approached 3 meters as the temperature approached 35 degrees and the width of the rectangle approached 4 meters as the temperature approached 35 degrees. Then it would be useful and sensible for me to conclude that the area of the rectangle would approach 12 square meters as the temperature approached 35 degrees.
• 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. •  This is an excellent question because it is a point that beginners are often confused about.

You are thinking of limits incorrectly. A limit is NOT what you would get if you actually did the math of the expression at the limiting value. The limit is what you would be approaching as you got extremely close to, but not equal to, the limiting value.

The whole point in bothering with limits is finding ways of getting values that you cannot directly compute (usually division by 0 or other undefined or indeterminate forms).

Thus, lim x→0 1/x² = infinity

You would not plug in x = 0, you would examine what happens when you get extremely close to x=0. For example, what is 1/x² when x = 1×10⁻¹²³? It is 10²⁴⁶. So, as you get closer and closer to x=0, clearly this is heading toward infinity.

Now, it is the case that IF and ONLY IF the expression is both defined and continuous at the limiting value, then the limit can be found just by plugging in the limiting value. However, if the expression is either not continuous or not defined at that point, then you must use other means of finding the limit.
• Where can I find the "rigorous proof" of these properties? • 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)? • 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? • 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? • 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 