7. Pinching Theorem

If a function is always trapped between two functions that approach the same limit at a point, then the trapped function must also approach that limit

$$\lim_{x \to a} f(x)$$

if we know that there exists two other functions $g(x)$ and $h(x)$ such that for all $x$ sufficiently close to $a$

$$g(x) \leq f(x) \leq h(x)$$

if the bounding functions have the same limit

$$\lim_{x \to a} g(x) = L, \quad \lim_{x \to a} f(x) = L$$

then $f(x)$ is forced to approach the same value, so

$$\lim_{x \to a} f(x) = L$$

 

Example

$$\lim_{x \to 0} x^2\sin(\dfrac{1}{x})$$

Answer

Let $t = \dfrac{1}{x}$ then for all real $t$

$$-1 \leq \sin(t) \leq 1$$

multiple all parts by $x^2$

$$-x^2 \leq x^2\sin(\dfrac{1}{x}) \leq x^2$$

take the limits of the bounding functions

$$\lim_{x \to 0} -x^2 = 0, \quad \lim_{x \to 0} x^2 = 0$$

since both bounds approach $0$ as $x \to 0$, the Pinching theorem implies

$$\lim_{x \to 0} x^2\sin(\dfrac{1}{x}) = 0$$