Motivating Questions
How can we use our knowedge of the limits of simple functions to determine the limits of more complicated functions?
How can we use our knowedge of the limits of simple functions to determine the limits of more complicated functions?
Let \(f(x)=\sin(1/x)\text{.}\)
The Squeeze Theorem (also known as the Sandwich Theorem) allows us to figure out some limits that we could not otherwise, by comparing them with known limits.
Evaluating the cell below illustrates that \(-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2\) (for \(x \ne 0\)). We cannot directly compute
because of the division by 0, but from the graph it is clear we should get 0.
The following theorem allows us to conclude that the limit really is 0.
If
then \(\displaystyle\lim_{x\rightarrow a} g(x) = L\,.\)
To apply the theorem to \(\displaystyle\lim_{x\rightarrow 0} x^2 \sin\left(\frac{1}{x}\right)\text{,}\) set
Since \(-1 \le \sin(\cdot) \le 1\) and \(0\le x^2\text{,}\) we have \(f(x)\le g(x) \le h(x)\) when \(x\) is near \(0\) (and for all \(x\neq 0\)). We can compute
so the assumptions of the Squeeze Theorem 1.1.1 are satisfied and we can conclude
Let \(g(x)=x \sin\left(\frac{1}{x}\right)\) and consider the limit \(\displaystyle\lim_{x\rightarrow 0} g(x)\text{,}\) which we cannot evaluate directly.
Use the Squeeze Theorem to find
Suppose we know \(f(x)\le g(x)\) when \(x\) is near \(a\) (except possibly at \(a\)) and \(\displaystyle\lim_{x\rightarrow a} f(x) = L\text{.}\) Can we conclude \(\displaystyle\lim_{x\rightarrow a} g(x) \ge L\text{?}\)