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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

# Theorem for limits of composite functions

Suppose we are looking for the limit of the composite function f(g(x)) at x=a. This limit would be equal to the value of f(L), where L is the limit of g(x) at x=a, under two conditions. First, that the limit of g(x) at x=a exists (and if so, let's say it equals L). Second, that f is continuous at x=L. If one of these conditions isn't met, we can't assume the limit is f(L).

## Want to join the conversation?

• Which "Limit property" allowing us to do this bracket "jump", I looked at the first video in this section and didn't figure it out. •   There's a confusing omission here in the video series, because (1) the "composition law" is assumed here but was never presented ("We can leverage our limit properties," the video says, but that property was left out); and (2) the extra condition for the composition law, namely that g(x) must be continuous at L, isn't mentioned. Nor do any of these examples bring up that extra condition. Instead, students are left to stumble over this condition by answering some of the following exercises wrong and then trying to figure out why they were wrong.

So, two recommendations: (a) Add the composition law to the "Limit properties" video, including the requirement for g(x) to be continuous at L; and (b) add to this video an example where the only thing stopping us from being able to evaluate the limit is that h(x) is not continuous at L. I.e. lim (x->L) h(x) must exist, but be different from h(L). Examples 2 and 3 in this video already fail to meet the extra condition, but they also fail to meet earlier conditions and so we don't get to see the continuity condition in action.
• how is lim g( f(x) ) can be changed to g( lim f(x) )? Khan please help us •  I couldn't find in the "Limit Properties" video mention of the rule which allows to decompose
``lim f(g(x))``
into
``f( lim g(x))``
, is it mentioned somewhere?
• At the starting how did the intuition of opening the bracket work?? • What Sal didn't mention in this video (but is explained later in the following exercise):
If functions `h(x)` and `g(x)` both have limits and and g(x) is continuous at `x=L`, [in other words ( `lim(x→c)h(x) = L` and `lim(x→c)g(x) = g(L)`] then
`lim(x→c) g(h(x)) = g(lim(x→c)h(x)) = g(L)`
So, if I'm not mistaken, since "outer" function `g(x)` should be continuous (in order for this property to hold) at the given limit then lim(x→c)g(L) = g(L), since function is defined at the limit, the limit is equal to function's output.
• Does Khanacademy have some videos on the proof of these properties ? • lim { x→1 } f ( g(x) ) = f ( lim { x→1 } g(x) ),
so, does that mean that this whole limit will exist as long as lim {x→1} g(x) is defined and f(that value of g) is defined, even if lim {x→g} f(x) does not exist?

For example, g(x) approaches 3 when x approaches 1, and f(3) = 10 but the function f(x) is discontinuous at f(3) such that the one side limits are different and hence its limit is undefined, will lim {x→1} f(g(x)) return the value 10? • No, and here is a counterexample. Define `g(x) = 3 - |x-1|`. Observe that `g(x) → 3` as `x → 1`. Define the function `ƒ` as follows: `ƒ(x) = 0` if `x < 3` and `ƒ(x) = 10` if `x ≥ 3`. For any `x ≠ 1`, one has `g(x) < 3`, and so `ƒ[g(x)] = 0` for such `x`. It follows that

`lim (x → 1) ƒ[g(x)] = lim (x → 1) 0 = 0 ≠ 10 = ƒ(3).`
• Hello, I think an example is missing.
Imagine lim (x -> -2) of f(g(x))
lim (x -> -2) g(x) is 0

f(0) = -1 but is not continuous such as lim ( x -> 0-) f(x) = 1 and lim ( x -> 0+) f(x) = -1

but f(0) is defined and equal -1 (in fact is just a discontinuity

what is the limit ? it is defined ?

thanks :) • What is the difference between "undefined" and "does not exist"? • At Sal evaluates the function at the point -3 is approaching the value of 1, but there is a discontinuity, a point that looks like it's at -3, 4. Why is the limit not 4?   