An interesting additional tool for evaluating limits:

Consider the limit \( \displaystyle \lim_{x \to 0} ~ \left\lvert x \right\rvert \sin{\left( \frac{\pi}{x} \right)} \). We cannot use the
limit product rule to evaluate the limit since \( \displaystyle \lim_{x \to 0} ~ \sin{\left( \frac{\pi}{x} \right)} \) does not exist.

Looking at the plot of the function on the interval \( \left[ -\frac{3}{2}, \frac{3}{2} \right] \)
notice that the oscillations grow increasingly more crowded as \( x \) approaches \( 0 \).

Zooming in on \( x = 0 \) by decreasing the width of the interval to \( [-1, 1] \) gives:

Decreasing the width of the interval to \( \left[ -\frac{1}{2}, \frac{1}{2} \right] \):

Decreasing the width of the interval to \( \left[ -\frac{1}{10}, \frac{1}{10} \right] \):

Observe that, no matter how small the viewing interval is, the function continues to oscillate extremely frequently as \( x \to \infty \). However, the amplitude of the oscillations appear to decrease linearly as \( x \to 0 \), as if the function values are bound, or “hemmed in” by lines of unit slope (observe that the amplitude approximately equals the magnitude of the \( x \) value, implying a slope of \( 1 \) on the bounding lines). This appears to force the limit to \( 0 \) as \( x \to 0 \).

This behaviour is made more apparent by combining the function with the apparent bounds \( \left\lvert x \right\rvert \) and \( -\left\lvert x \right\rvert \) on the intervals \( \left[ -1, 1 \right] \) (left) and \( \left[ -\frac{1}{10}, \frac{1}{10} \right] \) (right).

It is clear that the functions \( \left\lvert x \right\rvert \) and \( -\left\lvert x \right\rvert \) are acting to “funnel” the function \( \left\lvert x \right\rvert \sin{\left( \frac{\pi}{x} \right)} \) towards a limit of \( 0 \) as \( x \to 0 \). This is a result of the fact that the values of the sine function lie between \( 1 \) and \( -1 \) for any value of the argument; that is,

\( -1 \leq \sin{\left( \tfrac{\pi}{x} \right)} \leq 1 \) for all \( x \neq 0 \)

Thus, multiplying the inequality by \( \left\lvert x \right\rvert \) (which is non-negative), we see that

\( -\left\lvert x \right\rvert \leq \left\lvert x \right\rvert \sin{\left( \tfrac{\pi}{x} \right)} \leq \left\lvert x \right\rvert \) for all \( x \neq 0 \)

Since both of the “outer” functions \( \pm \left\lvert x \right\rvert \) approach \( 0 \) as \( x \to 0 \), so also does the “inner” function \( \left\lvert x \right\rvert \sin{\left( \frac{\pi}{x} \right)} \). Informally, the inner function gets “squeezed” or “sandwiched” between the two outer functions.

The above example illustrates the **squeeze theorem**, which essentially says that

If \( m(x)\), \(f(x)\), and \( M(x) \) are all defined near \( x = a \), and \[ m(x) \leq f(x) \leq M(x) \] near \( x = a \), with \[ \lim_{x \to a} ~ m(x) = L = \lim_{x \to a} ~ M(x) \] then, we also have \( \displaystyle \lim_{x \to a} ~ f(x) = L \).

- Use the squeeze theorem to show that \( \displaystyle \lim_{x \rightarrow 0} ~ x^2 \cos
\left(\frac{2}{x}\right) = 0 \).
- The squeeze theorem can also be applied as \( x \rightarrow \infty \). Given that
\[
\frac{3x-1}{x} \lt f(x) \lt 3 - \frac{1}{1+x^2} \text{for all } x \gt 10
\]
what is \( \displaystyle \lim_{x \rightarrow \infty} ~ f(x) \)?