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.

## AP®︎ Calculus AB (2017 edition)

### Unit 1: Lesson 7

Determining limits using direct substitution

# Limits by direct substitution

AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)
Sal explains how you can easily find limits of functions at points where the functions are continuous: simply plug in the x-value into the function! Later we will learn how to find limits even when the function isn't continuous.

## Want to join the conversation?

• I think Sal may have left something out. Correct me if I'm wrong, but on certain occasions, the entire function does not have to be continuous. If it is continuous at the value, a, then it should be fine to use direct substitution even though the function might be undefined elsewhere. This is because it isn't undefined at a so your output will be defined. Also, since your function is continuous at a, the limit there will exist meaning that you won't have to worry about jump discontinuities. Please tell me if this all seems reasonable.
• Because it is a continuous graph, every interval would be continuous provided it is a real number
• how would we know if the problem is to be solved by direct substitution method? or if the function given is continuous? without drawing its graph
• lim h(x) as x ->6 for h(x) =square root of (5x+6) should be 6 or -6 or both?
• y=√x is a function that returns only positive values. So in this context, we are only looking at the principal root, and the limit is positive 6.

Limits are unique: they cannot have multiple values. So "both" would never be an option.
• At , what's the difference between f(x) as x->a, and f(a)?
• One is a limit, the other is an evaluation of the function. If the function is continuous and defined at (in your example), a, then they're equivalent. But you can get some very interesting results if the function is not continuous or not defined.

The limit is basically saying what the function seems to be going to as x gets closer to closer to a, but the function may not be defined at that point.
• Is the limit zero or none since it is continuous?
• The limit is 0 as Sal has demonstrated in the video.
• Can a limit be a fraction?
• yes a limit can be a fraction. For example if I asked you to find the limit as X approaches 2 for the functions (1/X), when you use substitution to find the limit you find that the number it approaches is .5 or 1/2.
• I thought it like this : Lim x-> a f(x) = f(a) if f(x) is continuous at x=a, not vice versa? Can someone tell me why?
• That statement holds in both directions. A function is continuous at a point a if and only if the limit value equals the function value at a. Since the two statements are equivalent, this can be used as a definition of continuity.
• Ok so basically if a function, say f(x), is continuous at x=c, then the lim x-->c = f(x)? Is this why you can find limits of continuous functions by direct substitution?
• Yes, the limit as x->c of f(x) is f(c). This property is equivalent to the epsilon-delta definition of continuity, and it's why we can use direct substitution for most familiar functions.
• By Sal and almost all discussion about relation between continuity and limit: it always says that as long as the function is continuous, there is a limit.
i am wondering if there is a case that the function is continuous but the limit does not exist.

https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Active_Calculus_(Boelkins_et_al)/1%3A_Understanding_the_Derivative/1.7%3A_Limits%2C_Continuity%2C_and_Differentiability

in figure 1.7.2, the right graph, at point x=1. the left hand limit is 2. but the as x approaches 1 from the right, the function keeps oscillating, so the limit does not exist.

so the function is continuous by the graph, but limit does not exist.

am I wrong? or am I give some wrong example?
• Interesting question!

By definition, a function f(x) is continuous at x=x_0 if and only if for every epsilon>0, there exists delta>0 such that whenever x is within delta of x_0, f(x) is always within epsilon of f(x_0).

Though the function might look continuous at x=1, it is really discontinuous at x=1 due to the oscillatory behavior. For example, if we choose epsilon = 1/2, there does not exist delta>0 such that whenever x is within delta of 1, f(x) is always within epsilon of f(1). This is because there are values of x arbitrarily close to 1 (from the right) such that f(x) differs from f(1) by as much as 1 unit.
(1 vote)
• I'm not entirely sure if this statement is always true. But my argument bases on the assumption that the function f(x) = |x| is not continuous at x = 0, which I think is true but correct me if I'm wrong.

So, the limit lim_{x -> 0}f(x) for f(x) = |x| is definitely defined since it approaches 0 from both sides. The function itself is usually defined so that |x| equals 0 at x = 0. Which means that lim_{x -> a}f(x) equals f(a) at a = 0, but as said AFAIK the absolute value function |x| is NOT continuous for x = 0.

And that would disprove the statement that if lim_{x -> a}f(x) equaling f(a) will mean that f is continuous at a (and vice versa).
(1 vote)
• Hey Raphael! I think you're getting continuity and differentiation confused. For a function, f(x), to be continuous, f(c) = lim_{x->c}f(x). This means that f(c) = lim_{x->c-}f(x) = lim_{x->c+}f(x) = lim_{x->c}f(x). As long as this holds true the function will be continuous at the given x-value, c.

Differentiation on the other hand is different. Differentiation requires that a function is continuous at the given x-value along with being absent of cusps, sharp turns (like x=0 for f(x) = |x|), and vertical tangents.

So to answer your question, f(x) = |x| is continuous at x = 0, however, it is not differentiable at x = 0.