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### Course: Differential Calculus > Unit 1

Lesson 6: Limits by direct substitution- Limits by direct substitution
- Limits by direct substitution
- Undefined limits by direct substitution
- Direct substitution with limits that don't exist
- Limits of trigonometric functions
- Limits of trigonometric functions
- Limits of piecewise functions
- Limits of piecewise functions
- Limits of piecewise functions: absolute value

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# Limits of piecewise functions

In this video, we explore limits of piecewise functions using algebraic properties of limits and direct substitution. We learn that to find one-sided and two-sided limits, we need to consider the function definition for the specific interval we're approaching and substitute the value of x accordingly.

## Want to join the conversation?

- What does piecewise mean?(8 votes)
- It means we've defined the function as a bunch of separate curve segments on different intervals.(22 votes)

- At0:54is it not possible that the square root of 4 is also equal to -2, such that the limit of f(x) as x approaches 4 from the right does not exist? Or is the negative square root rejected?(6 votes)
- Hey there!

For your question, the negative square root is*not*rejected. It is simply because the functions asks for the*principal square root*; that is, the*positive*value of the square root. For your example, the function asks for the sqrt(x). This is interpreted as the principal square root of x. If we plug in 4, the principle square root would be positive 2.

For some additional context, if the function wanted both the positive*and*negative roots, we would write +/-sqrt(x), which if we plug in 4 would evaluate to both 2 and -2. Additionally, if we just want the negative roots, we could write -sqrt(x), which gives only -2 when we plug in 4.

Hope this clears some confusion!(14 votes)

- Hi, I am always wondering: What is the limit of the f(x) you give, while x is approaching 0? or it does not exist?(3 votes)
- One would use the appropriate one sided limit for such values at the endpoints of a domain. In this case the value approached by the function as x closes on 0 is, indeed, -2: lim x → 0+ = -2. However lim x → 0 does not exist because lim x → 0- does not exist as all values of x equal to or smaller than zero are not part of the domain of f(x).(13 votes)

- What background knowledge do I need to have before starting calculus? It will be nice to know what I need to know so I could learn this as clearly as possible.(2 votes)
- A calculus course will usually start from scratch with limits, so having previous experience with limits is helpful, but not strictly necessary.

You should be very comfortable with algebra and algebraic manipulations. Most calculus problems consist of many lines of algebra, and just a little calculus at the beginning or end. You should also be familiar with the trigonometric functions and have some basic trig identities easily available, like the Pythagorean and double-angle identities.

You'll need a grasp of series and sigma notation as well, and that should set you for Calculus 1 or AP Calculus AB.(14 votes)

- At3:15, why when x tends to -1 isn't the result 1/2? -1 IS in the second interval which would give us 2^-1= 1/2. I get why the limit doesn't exist though. I just find it weird that the -1 belongs to the interval.(3 votes)
- That's because limits don't care about what the function's actual value is at that point, as they only ask what value the function is approaching from both sides (which is two different values, hence the limit is undefined).(3 votes)

- For f(x), what will be the limit as x approaches 1?(2 votes)
- Good question, the limit does not exist though. The easiest way to tell is to graph it, you will see there is a vertical asymptote, where it heads toward infinity from the right and negative infinity from the left.

The only way a limit would exist is if there was something to "cancel out" the x-1 in the denominator. So if you had something like [(x+2)(x-1)]/(x-1). Then there would be a hole at 1, but the limit would still exist, and it would be 3. This is how you have to handle most rational functions.(2 votes)

- Sorry but can anyone give the intuition behind why anything to the power 0(a^0) is 1.

...also what about 0^0?(1 vote)- We know that increasing the exponent by 1 represents a multiplication by 𝑎 (the base):

𝑎^(𝑛 + 1) = (𝑎^𝑛) ∙ 𝑎

By the same logic,*decreasing*the exponent by 1 represents a*division*by 𝑎:

𝑎^(𝑛 − 1) = (𝑎^𝑛)∕𝑎

Thereby, we can write

𝑎^0 = 𝑎^(1 − 1) = (𝑎^1)∕𝑎 = 𝑎∕𝑎 = 1, as long as 𝑎 ≠ 0 because otherwise we get 0∕0 which is an indeterminate expression.(3 votes)

- At3:13Sal writes sin(-1+1)=0 . Shouldn't he have written (...)= 0+Pi*k?

Is there some rule that limits must have a single value?(1 vote)- There is a rule (and a theorem, not just a definition) that limits have at most one value, but it's not terribly relevant here. As the input of sin(x) approaches 0, sin(x) approaches 0 as well.

It's true that sin(x) approaches 0 around other x-values as well, but that's irrelevant here, because we're only looking at sin(x) around x=0 (or at sin(x+1) around x=-1).(2 votes)

- Before we leave tangent functions and limits... the graph of the tangent function at PI indicates the limit approaches -1. You are saying the limit approaches 0. Please advise the discrepancy.(1 vote)
- You are likely looking at the graph of cos(x), which does evaluate as (pi,-1). tan evaluates as (pi,0).(2 votes)

- I did an exercise on Khan Academy but I wasn't sure why. So the problem is the following:

g(x):

(first equation): (x/3)-2 for 0<x<6

(second equation): cos(x∙π) for 6≤x≤10

Then, it's asked to find the limit of g(x) when x is approaching 6.

So what's the answer for this question and why?(1 vote)- This shows that when you graph the function g(x) from 0 to 6 the function would be (x/3)-2 and from 6 to 10 the function would be cos(x*pi).

From the left side on the number line you can plug in 6 to the function:

(6/3) - 2 gives you 0

From the right side when you plug in 6 you get

cos(6 pi) which is equal to 1

Since the limit of g(x) is different from where the function is approaching from the right and the left the limit does not exist.

Hope this helps, if you need a clarification let me know.(2 votes)

## Video transcript

- [Instructor] Let's
think a little bit about limits of piecewise functions
that are defined algebraically like our f of x right over here. Pause this video and see
if you can figure out what these various limits would be, some of them are one-sided, and some of them are regular
limits, or two-sided limits. Alright, let's start with this first one, the limit as x approaches four, from values larger than equaling four, so that's what that plus tells us. And so when x is greater than four, our f of x is equal to square root of x. So as we are approaching
four from the right, we are really thinking about
this part of the function. And so this is going to be
equal to the square root of four, even though right at four, our f of x is equal to this, we are approaching from
values greater than four, we're approaching from
the right, so we would use this part of our function definition, and so this is going to be equal to two. Now what about our limit of f of x, as we approach four from the left? Well then we would use this
part of our function definition. And so this is going to
be equal to four plus two over four minus one, which is equal to 6 over three, which is equal to two. And so if we wanna say
what is the limit of f of x as x approaches four, well
this is a good scenario here because from both the left and the right as we approach x equals
four, we're approaching the same value, and we
know, that in order for the two side limit to have
a limit, you have to be approaching the same thing
from the right and the left. And we are, and so this is
going to be equal to two. Now what's the limit as x
approaches two of f of x? Well, as x approaches
two, we are going to be completely in this
scenario right over here. Now interesting things do
happen at x equals one here, our denominator goes to
zero, but at x equals two, this part of the curve
is gonna be continuous so we can just substitute
the value, it's going to be two plus two, over two minus
one, which is four over one, which is equal to four. Let's do another example. So we have another piecewise function, and so let's pause our video
and figure out these things. Alright, now let's do this together. So what's the limit as x
approaches negative one from the right? So if we're approaching from the right, when we are greater than
or equal to negative one, we are in this part of
our piecewise function, and so we would say, this
is going to approach, this is gonna be two, to
the negative one power, which is equal to one half. What about if we're
approaching from the left? Well, if we're approaching from the left, we're in this scenario right over here, we're to the left of
x equals negative one, and so this is going to
be equal to the sine, 'cause we're in this case,
for our piecewise function, of negative one plus one,
which is the sine of zero, which is equal to zero. Now what's the two-sided
limit as x approaches negative one of g of x? Well we're approaching
two different values as we approach from the right, and as we approach from the left. And if our one-sided
limits aren't approaching the same value, well then
this limit does not exist. Does not exist. And what's the limit of g of x, as x approaches zero from the right? Well, if we're talking
about approaching zero from the right, we are
going to be in this case right over here, zero is
definitely in this interval, and over this interval,
this right over here is going to be continuous,
and so we can just substitute x equals zero there, so it's
gonna be two to the zero, which is, indeed, equal
to one, and we're done.