Limits
8.0 Sandwich Theorem / Squeeze Play Theorem:
8.0 Sandwich Theorem / Squeeze Play Theorem:
Let $f(x)$, $g(x)$ and $h(x)$ be three real functions such that $f(x) \leqslant g(x) \leqslant h(x)$ for all $x$ in some interval $I$ containing the point $x=c$
and
if $\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} h(x) = L$, Then $\mathop {\lim }\limits_{x \to c} g(x) = L$
Question 22.
Evaluate $\mathop {\lim }\limits_{x \to 0} {x^3}\cos \frac{2}{x}$
Solution:
$ - 1 \leqslant \cos \frac{2}{x} \leqslant 1$,
$ - {x^3} \leqslant {x^3}\cos \frac{2}{x} \leqslant {x^3}$ for x>0
and ${x^3} \leqslant {x^3}\cos \frac{2}{x} \leqslant - {x^3}$ for x<0,
and $\mathop {\lim }\limits_{x \to 0} {x^3} = 0 = \mathop {\lim }\limits_{x \to 0} {( - x)^3}$
by sandwich theorem $\mathop {\lim }\limits_{x \to 0} {x^3}\cos \frac{2}{x} = 0$