Main content
Calculus, all content (2017 edition)
Squeeze theorem example
An example applying the squeeze theorem. Created by Sal Khan.
Want to join the conversation?
At2:12, when sal takes the limits , he uses the (greater or equal to inequality) sign. But shouldnt it be actually only equal sign , as the limits are equal as x approaches 2.(43 votes)
The limits are in fact equal, and it's easy enough to see that without resort to the squeeze theorem. The point of this exercise, though, is to show how the squeeze theorem could be used to establish this limit, so we use the inequality until the final step.(32 votes)
How do we know when we take the principal root (the positive root) or when it's the square root (negative and positive roots)?(11 votes)
Most often, context will tell you. But, if you are given the square root, it tends to be the principal square root; if you put it there yourself, it tends to be both square roots.(16 votes)
I just don't see any point in using the squeeze theorem. The limit could be easily found by factoring the expression in the numerator.(4 votes)- You are being given easy examples. There are times when you won't know how to find the limit of function itself, but can make use of squeeze.(24 votes)
Up till now i have watched almost every video of "limits" and the idea is pretty clear but what keeps me wondering is whether sal would talk about other ways of solving the limit functions rather then just using algebra or rationalising the functions. In school i was taught to differentiate the numerator and denominator of the function and put the value of x in it . For eg we need to find the limit of x^2-3x+2/x-2 ...after differentiating numerator and denominator we are getting 2x-3/1 , and by just putting the value of x in the function to get 1 . Why do we really need to differentiate ?(2 votes)
Your teachers were demonstrating a theorem called L' Hopitals Rule. In that case it might have not been so helpful but in other cases if you just plug it in you might get 0/0. In that case L'Hopital claimed that taking the limit of the derivative of the numerator/ the derivative of the denominator would result in the same limit.(6 votes)
- Wouldn't x confuse you with times?(2 votes)
Nope. It is a widespread convention to write times as a point, or as implicitly there when there is nothing. I mean when for example you see (5x+6y+4z); this equals [ 5*x+6*y+4*z]. Wouldn't it become pretty tricky to write these kinds of formulae with x's and y's, if we write the product with an x?(5 votes)
- isn't calculate the limit using algebra faster? Because with this theorem we first have to find the two functions that "sandwich" our function, then find proof that they are are indeed greater/less or equal than our function. Then we can really use the theorem. With algebra we can find the limit much faster(3 votes)
- You would normally use this theorem for those instances of a limit of which you don't know how to find the limit or for which the limit is too impractical to compute. So, you sandwich the difficult limit with two relatively easy limits.
So, if we could readily find the limit of the function, then we wouldn't bother with the squeeze theorem. So, it is ordinarily just used for the really hard to find limits.(2 votes)
When is this practical? I'm not challenging it, I believe it is, I'm just struggling to grasp when it can be used other than math-class-problems. When/how can you have a function, and easily know that it's less than or equal to some other function and greater than or equal to a second function? Especially without graphs to look at?(1 vote)
It is used primarily in the world of mathematics to help us understand the behavior of certain functions we didn't fully understand by using other functions which we did understand. In regards to their inequality, you can compare them by comparing inputs and outputs of the functions. Using your results, you could see which is greater.(5 votes)
If the three plotted fuctions meet at more than one point keeping the condition of squeeze theorem valid, periodically after a certain interval, then what to do?(2 votes)
As long as you can show that at the point for which you are trying to find the value, the requirements of the Squeeze Theorem holds, what the functions do elsewhere matters not.(1 vote)
- i have got the point when x approaches to 2 and we have the squeeze theorem.but what happens when we have x approaches to 3 ? then there is no squeeze happening. we have just a ineqality .(1 vote)
I thought this video was pretty clear. At each value of x, the functions f, g, an h are in order of magnitude: f (x) <= g (x) <= h (x). So, at x = 3, g is between f and h. As we approach x = 2, the functions all converge, and g is driven to the value of 1, between f's value of 1 and h's value of 1.(2 votes)
- 3:33also you could turn the second function into a form where you'd be able to plug in the 2 without making it indeterminate by factoring the top right?(1 vote)
- Yes, once you cancel out x-2 from both the top and the bottom, you are left with x-1 which when evaluated at 2 gives 1.(2 votes)
2:12
Video transcript
The graphs of f of x, g of x,
and h of x are shown below. Select and drag cards to
create a compound inequality that orders the
values of f of x, g of x, and h of x for x-values
near 2 but not at 2 itself. So for any of the x-values that
are depicted right over here, say, x is equal to 3, we see
that h of 3 is the largest, f of 3 is the smallest,
and g of 3 is in between. And that's true for
any of the x-values that they've depicted here. If we look at when x is equal
to 1, h of 1 is the largest, f of 1 is the smallest,
g of 1 is in between. So for all of the x-values
that they're depicting, f of x is less than or equal
to g of x, which is less than or equal to h of x. And the only place where they
equal based on this graph comes into play, it looks like
as we approach x equals 2, it looks like all the
functions are approaching 1. So that's where the equal
might come into play. But let's keep seeing what
they want us to do after that. So then it says it follows that. And instead of writing f
of x, g of x, and h of x, they've written the actual
definitions of them. So let's just remind
ourselves f of x is 2x times the square root of
x minus 1 minus 1. That's this blue
one right over here. So instead writing
f of x, we can write 2 times the square
root of x minus 1. Minus 1 is less than
or equal to g of x. G of x was this rational
expression right over here. So let's go back down here. We get this rational expression. And then that's going to
be less than or equal to h of x, which was, I believe,
e to the x minus 2. Is that right? Yep, e to the x minus 2. So all we've really
done is replaced f, g, and h with
their definitions. And then this means that
the limit-- so they're looking at the limit
as x approaches 2 of these different
expressions. So the limit as x approaches
2 of this expression is going to be less than
or equal to the limit as x approaches 2 of
this expression, which is this right over here,
which is going to be less than or equal to the
limit as x approaches 2 of this expression, which
is that right over there. And then they say
finally the value of the limit as x approaches 2
of this thing right over here is-- well, this is where
the squeeze theorem comes into play. We just have to remind
ourselves-- well, let's just think about it. Can we figure out the
limit as x approaches 2 of this right over here? Well, the limit as x approaches
2-- let's see, 2 minus 1. So we're taking the
principal root of 2 minus 1, which is the
principal root of 1. So you have 2 times 1 minus 1. So this is 1. This right over here
is e to the 2 minus 2. That's either the 0,
or that's 1 as well. So the limit of all
of this is going to be greater than
or equal to 1, and it's going to be
less than or equal to 1. Or it's right in
between 1 and 1. And the only way that it's
going to be between 1 and 1 is if it is equal to 1. This is the squeeze theorem
at play right over here. g of x, over the domain
that we've been looking at, or over the x-values
that we care about-- g of x was less than or equal
to h of x, which was-- or f of x was less than or equal
to g of x, which was less than or equal to h of x. And then we took the limit for
all of them as x approached 2. For the lower function, for
f of x, it approached 1. And we see it in the
graph right over here. In the lower function,
f of x approaches 1, h of x approaches 1,
and therefore, g of x must also approach 1. And we actually see that in
this graph right over here. But anyway, we could
check our answer just to feel good about ourselves. We got it right.