Question

In Exercises \(59 - 62 ,\) find the limit graphically. Use the Sandwich Theorem to confirm your answer. $$\lim _ { x \rightarrow 0 } x ^ { 2 } \cos \frac { 1 } { x ^ { 2 } }$$

Short answer

Using the graphical method and reconfirming with the Sandwich Theorem, the limit of \(x^{2} \cos(1/x^{2})\) as \(x \rightarrow 0\) is 0.

Step-by-step solution

Visualize the function graphically

Essentially, when x is very close to 0, the cosine function oscillates between -1 and 1 rapidly because of the term 1/\(x^{2}\) in the cosine. The function \(x^{2} \cos(1/x^{2})\) is the product of \(x^{2}\) (which is always non-negative) and \(\cos(1/x^{2})\), so it oscillates between -\(x^{2}\) and \(x^{2}\). The function may be difficult to graph due to rapid oscillations, but one should keep in mind the overall shape.

Apply the Sandwich Theorem

The sandwich theorem states that if \(g(x) \leq f(x) \leq h(x)\) for all x in some open interval containing a, except possibly at a itself, and if \(\lim_{x \rightarrow a}g(x) = \lim_{x \rightarrow a}h(x) = L,\) then \(\lim_{x \rightarrow a}f(x) = L.\) Now in this case, when \(x \neq 0\), we have \(- | x | ^ { 2 } \leq x ^ { 2 } \cos ( 1 / x ^ { 2 } ) \leq | x | ^ { 2 } = x^2.\) As \(x \rightarrow 0, x^2\) and \(-x^2\) also approach 0, so the Sandwich Theorem implies that \(\lim _ { x \rightarrow 0 } x ^ { 2 } \cos ( 1 / x ^ { 2 } ) = 0.\)

Confirm with the Graph

On a graph, \(x^{2}\) and \(-x^{2}\) are the upper and lower bounds of the function \(x^{2} \cos(1/x^{2})\). Both of these bounds converge to 0 as \(x \rightarrow 0\), reconfirming the finding that \(\lim _ { x \rightarrow 0 } x ^ { 2 } \cos ( 1 / x ^ { 2 } ) = 0.\)