# Squeeze Principle

### Squeeze Principle

Assume that function f.g and h satisfy g(x) ≤ f (x) ≤ h (x) and $$\underset{x\to a}{\mathop{\lim }}\,g(x)=L=\underset{x\to a}{\mathop{\lim }}\,h(x)$$ then $$\underset{x\to a}{\mathop{\lim }}\,h(x)=L$$.

Note: The quantity a may be a finite number +∞ or -∞. The quantity L may be finite number +∞ or -∞.

The squeeze principle is used on limit problems where the useful algebraic methods (factoring conjugation, algebraic manipulation, etc..) are not effective. However, it requires that you be able to ‘squeeze’ your problem in between two other simpler functions where limits are easily computable and equal.

Example: Compute $$\underset{x\to \infty }{\mathop{\lim }}\,\frac{sin x}{x}$$

Solution: First note that

-1 ≤ sin x ≤ 1 thus, -1/x ≤ sin x/ x ≤ 1/x

Since, $$\underset{x\to \infty }{\mathop{\lim }}\,\frac{-1}{x}=0=\underset{x\to \infty }{\mathop{\lim }}\,\frac{1}{x}$$

It follows from the squeeze principle that $$\underset{x\to \infty }{\mathop{\lim }}\,\frac{sin x}{x}=0$$

Example: Compute $$\underset{x\to {{o}^{-}}}{\mathop{\lim }}\,{{x}^{3}}\cos \left( \frac{2}{x} \right)$$.

Solution: Note that $$\underset{x\to {{o}^{-}}}{\mathop{\lim }}\,{{x}^{3}}\cos \left( \frac{2}{x} \right)$$. does not exist since values of cos (2/x) oscillate between – 1 and +1 ass x approaches o from the left. However this does not necessarily mean that $$\underset{x\to {{o}^{-}}}{\mathop{\lim }}\,{{x}^{3}}\cos \left( \frac{2}{x} \right)=0$$ does not exist. Indeed x³ < 0 and-1 ≤ cos (2/x) ≤ +1 for x < 0 multiply each component by x³ reversing the inequalities and getting -x³ ≥ x³ cos (2/x) ≥ x³.

Or, x³ ≤ cos (2/x) ≤ – x³.

Since, $$\underset{x\to {{o}^{-}}}{\mathop{\lim }}\,{{x}^{3}}=0=\underset{x\to {{o}^{-}}}{\mathop{\lim }}\,\left( -{{x}^{3}} \right)$$,

∴ Follows from the squeeze principle that $$\underset{x\to {{o}^{-}}}{\mathop{\lim }}\,{{x}^{3}}\cos \left( \frac{2}{x} \right)=0$$.