Differential Calculus From Practice to Theory

It is clear from the graph below that \(\limit{x}{\infty}{\sin(x)}\) does not exist since the \(\sin(x)\) has no horizontal asymptotes. To say that this limit does not exist means that \(y=\sin(x)\) is not approaching any particular value as \(x\rightarrow\infty\text{.}\) It’s not even “approaching” \(+\infty\) or \(-\infty\) since it is bounded both above and below.

We need a new idea.

Look at the graph of \(f(x) = \frac{\sin(x)}{x}\text{:}\)

Can you see that the horizontal asymptote is \(y=0\text{?}\)

We will analyze what happens when \(x\rightarrow\infty\) and leave the case \(x\rightarrow-\infty\) for you in Problem 12.2.1.

We know that both \(\pm\frac1x\rightarrow0\) as \(x\rightarrow\infty\text{.}\) Since \(\frac{\sin(x)}{x}\) is caught between \(\frac1x\) and \(-\frac1x\) it follows that, for positive values of \(x\text{,}\) the highest and lowest points on the graph of \(\frac{\sin(x)}{x}\) must also approach zero as \(x \rightarrow\infty\text{.}\) Therefore \(\tlimit{x}{\infty}{\frac{\sin(x)}{x}}=0\text{.}\)

This idea is often called the Squeeze Theorem or the Sandwich Theorem.

In the previous example, notice that we had \(\frac{-1}{x}\le \frac{\sin(x)}{x}\le\frac1x\) for all \(x\neq0\text{,}\) but this is more than is actually required. We only need for the inequality to be satisfied when \(x\) sufficiently large, in either the positive or negative direction. That’s why we only require that \(\alpha(x)\le f(x)\le \beta(x)\) on the interval \((c,\infty)\) (positive case) or \((-\infty,c)\) (negative case) rather than for all values of \(x\text{.}\)