Squeeze theorem applied to an exponential-trig limit

Since $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ holds for every $x \neq 0$, and $e^x > 0$ for all real $x$, multiplying every part of this inequality by $e^x$ preserves the direction of the inequalities: $-e^x \le e^x \sin\left(\frac{1}{x}\right) \le e^x.$ This gives us bounding functions $-e^x$ and $e^x$ that sandwich $g(x)$ near $x=0$.

Now we evaluate the limits of these bounding functions as $x \to 0$. Since both $-e^x$ and $e^x$ are continuous at $x=0$, we use direct substitution: $\lim_{x \to 0} (-e^x) = -e^0 = -1$ and $\lim_{x \to 0} e^x = e^0 = 1.$ Crucially, these two limits are different: $-1 \neq 1$.

The Squeeze Theorem requires that the upper and lower bounding functions approach the same finite value. Since our bounding functions approach different limits, the conditions of the Squeeze Theorem are not satisfied, and we cannot use it to conclude that $g(x)$ has a limit.

To determine whether the original limit exists, we analyze the behavior directly. As $x \to 0$, we have $e^x \to 1$, while $\sin(1/x)$ oscillates between $-1$ and $1$ infinitely many times. Therefore, $g(x) = e^x \sin(1/x)$ oscillates between values near $-1$ and $1$ without converging to any single value.

Conclusion: $\lim_{x \to 0} e^x \sin\left(\frac{1}{x}\right)$ does not exist (DNE). This problem illustrates an important limitation: the Squeeze Theorem cannot always be applied, even when bounding functions are available. The theorem requires equal limits at the boundary.