Squeeze theorem

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  1. intuition behind the "Squeeze Theorem"…
  1. Prove that limx0  x10cos3πx=0lim_{x \to 0} \;{x^{10}}\cos \frac{{3\pi }}{x} = 0
    1. If 5g(x)4x3+9x2x15 \le g\left( x \right) \le 4{x^3} + 9{x^2} - x - 1, find limx2  g(x)lim_{x \to - 2} \;g\left( x \right)
      Topic Notes
      In this section, we will learn about the intuition and application of the squeeze theorem (also known as the sandwich theorem). We will recognize that by comparing a function with other functions which we are capable of solving, we can evaluate the limit of a function that we couldn't solve otherwise, using the algebraic manipulation techniques covered in previous sections.

      So far, we have looked at how to both find and evaluate limits for simple and more complex functions. However, as most topics in mathematics go, there will be times when evaluating limits will become impossible without using other techniques. That is where "Squeeze Theorem" comes in handy.

      What is the Squeeze Theorem

      Before we get into the mathematical Squeeze Theorem definition, let's first think of the concept in more familiar terms. Imagine I am a runner along with two of my friends, John and David. Instead of knowing how far I run each time, I know my distance compared to John and David according to the following:

      1) I always run equal to or further than John.

      2) I always run less than or equal to David.

      Let's say that one day myself, John, and David all go out running. On this particular day, John and David both run 3km. Now, the question is, how far did I run? If we answer this question according to the two conditions stated above, we will arrive at the only possible answer – 3km. Now let's break this down algebraically so you can see this clearly:

      - Let the distance I run be represented by M(x)

      - Let the distance John runs be represented by J(x)

      - Let the distance David runs be represented by D(x)

      Now, according to the two conditions above:

      J(x)M(x)D(x)J(x) \leq M(x) \leq D(x)

      And if both John and David ran 3km each:

      3M(x)33 \leq M(x) \leq 3

      This leads us to our answer of 3km – which is the only value that can satisfy this above equality. It was the functions that described the distance ran by both John and David that "squeezed" or "sandwiched" my distance function, and allowed us to find the answer without knowing a thing about the distance I ran. This is the idea behind "Squeeze" or "Sandwich" Theorem – it allows us to calculate the limit of a function using two other, more simple functions, when other methods aren't useful. For a more algebraic-based Squeeze Theorem proof, if you're interested, look here.

      How to Use Squeeze Theorem

      In mathematical terms, Squeeze Theorem is defined by the following:

      g(x)f(x)h(x)g(x) \leq f(x) \leq h(x)

      limnA  g(x)=lim_{n \to A} \;g\left( x \right)= limnA  h(x)=Llim_{n \to A} \;h\left( x \right)=L

      Therefore: limnA  f(x)=Llim_{n \to A} \;f\left( x \right)=L

      Basically, for a given inequality of three terms such as the one above – the limit of f(x) term is equal to the limit of g(x) and h(x) if those limits are equal to each other. Thereby, if we can evaluate the limit of both g(x) and h(x), and they equal each other, we can easily find the limit of f(x). Now, before we look at some concrete examples of how and when to do squeeze theorem, let's first review how to find and evaluate limits.

      How to Evaluate Limits

      Finding limits may seem like an intimidating process, especially when we are dealing the concept of infinity. The best way to review the process of how to find the limit of a function is to do an example problem:


      Find: limx  16x34x7+200lim_{x \to \infty } \;\frac{{{16x^3} - 4}}{{x^7 + 200}}

      Step 1: Eliminate Unnecessary Constants

      Because we are subbing in infinity for x (not actually – but we'll get to that later) the first step is to ignore the constants, as they will not change during this substitution. This makes our expression much more simple and easy to work with when we try to find the limit.

      limx  16x3x7lim_{x \to \infty } \;\frac{{{16x^3} }}{{x^7}}

      Step 2: Evaluate the Limit

      Now that we have simplified our expression, we can now evaluate the limit. Since infinity is not actually a number, we can't substitute it in for x – as alluded to in the previous step. What we can do, however, is imagine how our expression will change as we substitute larger and larger numbers in for x. This can be done by comparing the degrees of the numerator and denominator of our expression. Let's look at our expression again:

      limx  16x3x7lim_{x \to \infty } \;\frac{{{16x^3} }}{{x^7}}

      As you can see, our numerator has a degree of 3, and our denominator has a degree of 7. Thus, since the degree is higher on the denominator, our denominator will quickly outpace the growth of our numerator. Therefore, as x gets bigger and bigger, our numerator will essentially become 0 compared to our denominator, and thus the limit of this function as x approaches infinity is 0.

      limx  16x3x7lim_{x \to \infty } \;\frac{{{16x^3} }}{{x^7}} =0x7=0= \frac{0}{x^7} = 0

      There! Now that we have reviewed how to find the limit, let's get back to how to do Squeeze Theorem.

      How to Do Squeeze Theorem

      Though Squeeze Theorem can theoretically be used on any set of functions that satisfy the above conditions, it is particularly useful when dealing with sinusoidal functions. As with most things in mathematics, the best way to illustrate how to do Squeeze Theorem is to do some Squeeze Theorem problems.

      Example 1:

      Find limx  cosxxlim_{x \to \infty } \;\frac{{{\cos x} }}{{x}}

      Before we get into solving this problem, let's first consider why using Squeeze Theorem is necessary in this case. The simple answer to this is that we cannot possibly compute the cos of infinity. Because infinity is not actually a number, there is nothing we can substitute in for x in order to find this limit. For this reason, we must use Squeeze Theorem.

      Step 1: Make an Inequality

      Because of the nature of the cosine function: 1cosx1-1 \leq \cos x \leq 1

      Step 2: Modify the Inequality

      You'll notice that our inequality is insufficient for solving this problem, as we are asked to find the limit of cosx/x, not cosx alone. So we need to make some modifications. Dividing the entire inequality by x gives us what we are asked to solve.

      1xcosxx1x-\frac{1}{x} \leq \frac{\cos x}{x} \leq \frac{1}{x}

      Step 3: Evaluate the Left and Right Hand Limits

      Remember – thought the problem has asked us to solve the limit of cosx/x for when x moves to infinity, we will do so by finding the limit of the other two functions we have created in the inequality.

      limx  1x=lim_{x \to \infty } \;-\frac{1}{x} = limx  1x=0lim_{x \to \infty } \;\frac{1}{x}=0

      NOTE: If the left and right hand limits do not equal each other, we cannot utilize Squeeze Theorem.

      Step 4: Apply the Squeeze Principle

      Since we have now found the limit of both the left and right hand equations from our inequality, and they are equal to each other, we can use Squeeze Theorem to determine:

      limx  cosxx=0lim_{x \to \infty } \;\frac{\cos x}{x} =0

      That's it! We've successfully used Squeeze Theorem to find the limit of this function. Let's try some more complicated examples.

      Example 2:

      Find limx  x(sinx+cos3x)(x4+2)(x7)lim_{x \to \infty } \;\frac{x(\sin x+ \cos^3 x)}{(x^4+2)(x-7)}

      Step 1: Make an Inequality

      Because of the nature of the cosine function: 1cosx1-1 \leq \cos x \leq 1

      Step 2: Modify the Inequality

      Again, you'll notice that our inequality is insufficient for solving this problem, so we need to make quite a few modifications.

      Cubing the inequality: 1cos3x1-1 \leq \cos^3 x \leq 1

      Add the inequality 1sinx1-1 \leq \sin x \leq 1: 2sinx+cos3x2-2 \leq \sin x+ \cos^3 x \leq 2

      Multiply by x: 2xx(sinx+cos3x)2x-2x \leq x(\sin x+ \cos^3 x) \leq 2x

      Divide by (x4+2)(x^4+2): 2x(x4+2)x(sinx+cos3x)(x4+2)2x(x4+2)\frac{-2x}{(x^4+2)} \leq \frac{x(\sin x + \cos^3 x)}{(x^4+2)} \leq \frac{2x}{(x^4+2)}

      Divide by (x7)(x-7): 2x(x4+2)(x7)x(sinx+cos3x)(x4+2)(x7)2x(x4+2)(x7)\frac{-2x}{(x^4+2)(x-7)} \leq \frac{x(\sin x + \cos^3 x)}{(x^4+2)(x-7)} \leq \frac{2x}{(x^4+2)(x-7)}

      Now we have successfully constructed the function we were asked to find the limit of!

      Step 3: Evaluate the Left and Right Hand Limits

      Since these functions are much more complex than in the previous example, let's evaluate the left and right hand limits individually.

      Left: limx  2x(x4+2)(x7)=0lim_{x \to \infty } \;\frac{-2x}{(x^4+2)(x-7)}=0

      Using what we know about evaluating limits, hopefully you'll notice that as x gets bigger and bigger, our denominator will quickly grow much bigger than our numerator, as x5x^5 >> xx. This will bring our function closer and closer to zero.

      Right: limx  2x(x4+2)(x7)=0lim_{x \to \infty } \;\frac{-2x}{(x^4+2)(x-7)}=0

      The same logic can be used to find that the right hand limit also goes to zero.

      Step 4: Apply the Squeeze Principle

      Since we have again found the limit of both the left and right hand equations from our inequality, and they are equal to each other, we can use Squeeze Theorem to determine:

      limx  x(sinx+cos3x)(x4+2)(x7)=0lim_{x \to \infty } \;\frac{x(\sin x+ \cos^3 x)}{(x^4+2)(x-7)}=0