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1.8 Determining Limits Using the Squeeze Theorem

7 min readβ€’january 9, 2023

Anusha Tekumulla

Anusha Tekumulla

ethan_bilderbeek

ethan_bilderbeek


AP Calculus AB/BC ♾️

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The Squeeze Theorem πŸ₯ͺ

The Squeeze Theorem, also known as the Sandwich Theorem or the Squeeze Lemma, is a useful tool for determining the limit of a function as it approaches a particular value. It states that if the values of two functions, f(x) and g(x), are always bounded by the values of a third function, h(x), for all values of x within a certain range, then the limit of f(x) as x approaches a particular value must also be bounded by the limits of h(x) as x approaches the same value.
To use the Squeeze Theorem, we must first identify two functions, f(x) and g(x), that are bounded by a third function, h(x). For example, let's consider the function f(x) = sin(x), g(x) = x, and h(x) = x^2. We can see that for all values of x, the value of f(x) is always between the values of g(x) and h(x). That is, for all x:
g(x) ≀ f(x) ≀ h(x)
Now, let's consider the limit of f(x) as x approaches 0. We can use the Squeeze Theorem to show that this limit must be bounded by the limits of g(x) and h(x) as x approaches 0.
To do this, we must first find the limits of g(x) and h(x) as x approaches 0. The limit of g(x) as x approaches 0 is simply 0, since g(x) = x and x approaches 0 as x approaches 0. The limit of h(x) as x approaches 0 is also 0, since h(x) = x^2 and x^2 approaches 0 as x approaches 0.
Now that we have found the limits of g(x) and h(x) as x approaches 0, we can use the Squeeze Theorem to show that the limit of f(x) as x approaches 0 must also be between these two limits. That is, the limit of f(x) as x approaches 0 is bounded by 0 ≀ f(x) ≀ 0.
Using the Squeeze Theorem, we can conclude that the limit of f(x) as x approaches 0 is 0.
The Squeeze Theorem can also be used to show that the limit of a function does not exist. For example, consider the function f(x) = x^2 sin(1/x). We can see that as x approaches 0, the values of f(x) oscillate between positive and negative infinity. This means that there is no single value that f(x) approaches as x approaches 0, and therefore the limit of f(x) as x approaches 0 does not exist.
The Squeeze Theorem is a powerful tool for determining the limit of a function as it approaches a particular value. By finding two functions that are bounded by a third function, we can use the Squeeze Theorem to show that the limit of the first function is bounded by the limits of the other two functions. This can be used to determine the value of the limit, or to show that the limit does not exist.
Here are a few examples of using the Squeeze Theorem to determine the limit of a function:
Example 1:
Find the limit of f(x) = x^3 as x approaches 0.
We can use the Squeeze Theorem to show that the limit of f(x) as x approaches 0 is 0. To do this, we need to find two functions, g(x) and h(x), that are bounded by f(x) for all values of x within a certain range.
One possible choice for g(x) is g(x) = 0, which is always less than or equal to f(x) for all values of x. Another possible choice for h(x) is h(x) = x^2, which is always greater than or equal to f(x) for all values of x.
Therefore, we can use the Squeeze Theorem to conclude that the limit of f(x) as x approaches 0 is bounded by the limits of g(x) and h(x) as x approaches 0. That is, 0 ≀ f(x) ≀ 0.
Since both of these limits are equal to 0, we can conclude that the limit of f(x) as x approaches 0 is also 0.
Example 2:
Find the limit of f(x) = x^2 sin(1/x) as x approaches 0.
In this case, we can see that as x approaches 0, the values of f(x) oscillate between positive and negative infinity. This means that there is no single value that f(x) approaches as x approaches 0, and therefore the limit of f(x) as x approaches 0 does not exist.
Example 3:
Find the limit of f(x) = (x^2 - 4)/(x - 2) as x approaches 2.
We can use the Squeeze Theorem to show that the limit of f(x) as x approaches 2 is 3. To do this, we need to find two functions, g(x) and h(x), that are bounded by f(x) for all values of x within a certain range.
One possible choice for g(x) is g(x) = x + 1, which is always less than or equal to f(x) for all values of x. Another possible choice for h(x) is h(x) = x, which is always greater than or equal to f(x) for all values of x.
Therefore, we can use the Squeeze Theorem to conclude that the limit of f(x) as x approaches 2 is bounded by the limits of g(x) and h(x) as x approaches 2. That is, 2 ≀ f(x) ≀ 3.
Since both of these limits are equal to 3, we can conclude that the limit of f(x) as x approaches 2 is also 3.
Example 4:
Find the limit of f(x) = (x^3 - 27)/(x - 3) as x approaches 3.
We can use the Squeeze Theorem to show that the limit of f(x) as x approaches 3 is 0. To do this, we need to find two functions, g(x) and h(x), that are bounded by f(x) for all values of x within a certain range.
One possible choice for g(x) is g(x) = 0, which is always less than or equal to f(x) for all values of x. Another possible choice for h(x) is h(x) = x - 3, which is always greater than or equal to f(x) for all values of x.
Therefore, we can use the Squeeze Theorem to conclude that the limit of f(x) as x approaches 3 is bounded by the limits of g(x) and h(x) as x approaches 3. That is, 0 ≀ f(x) ≀ 0.
Since both of these limits are equal to 0, we can conclude that the limit of f(x) as x approaches 3 is also 0.
Example 5:
Find the limit of f(x) = (x^2 + 1)/(x^2 - 1) as x approaches 1.
We can use the Squeeze Theorem to show that the limit of f(x) as x approaches 1 is 2. To do this, we need to find two functions, g(x) and h(x), that are bounded by f(x) for all values of x within a certain range.
One possible choice for g(x) is g(x) = x + 1, which is always less than or equal to f(x) for all values of x. Another possible choice for h(x) is h(x) = x, which is always greater than or equal to f(x) for all values of x.
Therefore, we can use the Squeeze Theorem to conclude that the limit of f(x) as x approaches 1 is bounded by the limits of g(x) and h(x) as x approaches 1. That is, 1 ≀ f(x) ≀ 2.
Since both of these limits are equal to 2, we can conclude that the limit of f(x) as x approaches 1 is also 2.
Example 6:
Find the limit of f(x) = (x^3 - 8)/(x - 2) as x approaches 2. To complete this example using the Squeeze Theorem, we need to find the limits of g(x) and h(x) as x approaches 2. The limit of g(x) as x approaches 2 is 4, since g(x) = x^2 and x^2 approaches 4 as x approaches 2. The limit of h(x) as x approaches 2 is also 4, since h(x) = x + 2 and x + 2 approaches 4 as x approaches 2.
Now that we have found the limits of g(x) and h(x) as x approaches 2, we can use the Squeeze Theorem to show that the limit of f(x) as x approaches 2 is bounded by these limits. That is, 4 ≀ f(x) ≀ 4.
Since both of these limits are equal to 4, we can conclude that the limit of f(x) as x approaches 2 is also 4.
Therefore, the limit of f(x) as x approaches 2 is 4.
In conclusion, the Squeeze Theorem is a useful tool for determining the limit of a function as it approaches a particular value. It states that if the values of two functions, f(x) and g(x), are always bounded by the values of a third function, h(x), for all values of x within a certain range, then the limit of f(x) as x approaches a particular value must also be bounded by the limits of h(x) as x approaches the same value. By finding two functions that are bounded by a third function, we can use the Squeeze Theorem to determine the value of the limit or show that the limit does not exist. Understanding and applying the Squeeze Theorem can be a valuable skill in solving limits in AP Calculus and beyond.
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