Problems based on Sandwich Theorem
Problems based on Sandwich Theorem
The Sandwich Theorem, also known as the Squeeze Theorem or the Pinching Theorem, is a fundamental result in calculus that allows us to determine the limit of a function based on the limits of two other functions that "sandwich" it. This theorem is particularly useful when dealing with limits that are not straightforward to compute directly.
Understanding the Sandwich Theorem
The Sandwich Theorem states that if $f(x)$, $g(x)$, and $h(x)$ are functions such that $f(x) \leq g(x) \leq h(x)$ for all $x$ in some interval around $a$, except possibly at $a$ itself, and
$$ \lim_{{x \to a}} f(x) = \lim_{{x \to a}} h(x) = L, $$
then
$$ \lim_{{x \to a}} g(x) = L. $$
In other words, if $g(x)$ is "sandwiched" between $f(x)$ and $h(x)$, and both $f(x)$ and $h(x)$ approach the same limit $L$ as $x$ approaches $a$, then $g(x)$ must also approach $L$.
Important Points and Differences
| Feature | Description |
|---|---|
| Applicability | The Sandwich Theorem is used when the limit of $g(x)$ is difficult to find directly, but we can find two functions $f(x)$ and $h(x)$ that are easier to work with. |
| Conditions | The functions $f(x)$ and $h(x)$ must converge to the same limit at $a$, and $g(x)$ must be bounded by $f(x)$ and $h(x)$ near $a$. |
| Interval | The inequality $f(x) \leq g(x) \leq h(x)$ must hold in some interval around $a$, not necessarily for all $x$. |
| Exceptions | The inequality does not need to hold at the point $a$ itself, which is useful when dealing with functions that are undefined at $a$. |
Examples
Example 1: Basic Application
Suppose we want to find the limit of $g(x) = x^2 \sin(\frac{1}{x})$ as $x$ approaches 0. Direct computation is difficult due to the $\sin(\frac{1}{x})$ term. However, we know that $-1 \leq \sin(\frac{1}{x}) \leq 1$ for all $x \neq 0$. We can use this to "sandwich" $g(x)$:
$$ -x^2 \leq x^2 \sin(\frac{1}{x}) \leq x^2. $$
Now, we compute the limits of the bounding functions as $x$ approaches 0:
$$ \lim_{{x \to 0}} (-x^2) = \lim_{{x \to 0}} x^2 = 0. $$
By the Sandwich Theorem, we can conclude that:
$$ \lim_{{x \to 0}} x^2 \sin(\frac{1}{x}) = 0. $$
Example 2: Trigonometric Functions
Consider the function $g(x) = x \cos(x)$ as $x$ approaches 0. We can bound this function using the fact that $|\cos(x)| \leq 1$:
$$ -x \leq x \cos(x) \leq x. $$
Both bounding functions approach 0 as $x$ approaches 0:
$$ \lim_{{x \to 0}} (-x) = \lim_{{x \to 0}} x = 0. $$
Therefore, by the Sandwich Theorem:
$$ \lim_{{x \to 0}} x \cos(x) = 0. $$
Example 3: Combining Functions
Let's find the limit of $g(x) = x^3 \sin(\frac{1}{x})$ as $x$ approaches 0. Again, we use the fact that $|\sin(\frac{1}{x})| \leq 1$:
$$ -x^3 \leq x^3 \sin(\frac{1}{x}) \leq x^3. $$
The limits of the bounding functions are:
$$ \lim_{{x \to 0}} (-x^3) = \lim_{{x \to 0}} x^3 = 0. $$
Applying the Sandwich Theorem gives us:
$$ \lim_{{x \to 0}} x^3 \sin(\frac{1}{x}) = 0. $$
Conclusion
The Sandwich Theorem is a powerful tool in calculus for evaluating limits that are otherwise difficult to handle. By finding appropriate bounding functions that are easier to work with, we can determine the behavior of a more complex function as it approaches a particular point. Remember that the key to using the Sandwich Theorem effectively is to ensure that the bounding functions converge to the same limit and that the function of interest is properly "sandwiched" between them near the point of interest.