← Back to learning objective

Squeeze theorem applied to an oscillating exponential factor

Since −1≤sin⁡(1/x)≤1-1 \le \sin(1/x) \le 1 for all x≠0x \neq 0, and the exponential function is increasing, we have e−1≤esin⁡(1/x)≤e1e^{-1} \le e^{\sin(1/x)} \le e^{1} for all x≠0x \neq 0. Multiplying through by x2≥0x^2 \geq 0 gives the bound x2e≤x2esin⁡(1/x)≤x2e\dfrac{x^2}{e} \le x^2 e^{\sin(1/x)} \le x^2 e for all x≠0x \neq 0.

Using the bound given above, evaluate lim⁡x→0x2esin⁡(1/x).\lim_{x \to 0} x^2 e^{\sin(1/x)}.

Hint

Compute the limits of the lower bound x2/ex^2/e and the upper bound x2ex^2 e as x→0x \to 0 separately. If both bounding functions approach the same value, the Squeeze Theorem tells you the function trapped between them must approach that same value.

Step-by-step solution

We are given that x2e≤x2esin⁡(1/x)≤x2e\dfrac{x^2}{e} \le x^2 e^{\sin(1/x)} \le x^2 e for all x≠0x \neq 0. Since both bounding functions are simple polynomials times a constant, direct substitution applies: lim⁡x→0x2e=0e=0\lim_{x\to 0} \dfrac{x^2}{e} = \dfrac{0}{e} = 0 and lim⁡x→0x2e=0⋅e=0\lim_{x\to 0} x^2 e = 0 \cdot e = 0.

Both the lower and upper bounding functions tend to the same value, 00, as x→0x \to 0. By the Squeeze Theorem, the function trapped between them, f(x)=x2esin⁡(1/x)f(x) = x^2 e^{\sin(1/x)}, must also tend to that common value. Therefore lim⁡x→0x2esin⁡(1/x)=0\lim_{x \to 0} x^2 e^{\sin(1/x)} = 0.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap