Next Lesson

Example 1: Squeeze Theorem

Unit 1: Limits and Continuity Option 11: Squeeze Theorem (Sandwich Theorem) Example 1: Squeeze Theorem

It can be show that the inequalities

$1 - \frac{x^{2}}{6} < \frac{x \cdot \sin ((x))}{2 - 2 \cdot \cos ((x))} < 1$

hold for all values of x close to zero. What, if anything, does this tell you about

$\lim_{x \to 0} \frac{x \cdot \sin ((x))}{2 - 2 \cos ((x))} ?$

Step 1: Try Option 1: Plug it in

Always try plugging in the $x - value$ you are heading towards.

In this example we get division by zero, which means Option 1 has failed.

You could try all of the other limit options, but you will see really quick that none of those is going to work either.

$\lim_{x \to 0} \frac{x \cdot \sin ((x))}{2 - 2 \cos ((x))} = \frac{0 \cdot \sin ((0))}{2 - 2 \cos ((0))} = \frac{0}{0}$

Step 2: Confirm that you have three equations that fit the Squeeze Theorem Inequality, $g (x) \leq f (x) \leq h (x)$ .

In this example you are given this is true to start with. That is what the “strange” language at the beginning of the problem is telling you.

It can be show that the inequalities

$1 - \frac{x^{2}}{6} < \frac{x \cdot \sin ((x))}{2 - 2 \cdot \cos ((x))} < 1$

hold for all values of x close to zero.

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

$1 - \frac{x^{2}}{6} < \frac{x \cdot \sin ((x))}{2 - 2 \cdot \cos ((x))} < 1$

Step 3: Apply the limit portion of the Squeeze Theorem at the given x-value = c , to the outside equations to confirm $\lim_{x \to c} g (x) = \lim_{x \to c} h (x) = L$ .

Here the c-value that you will use is c = 0 since the final limit you want an answer to,
$\lim_{x \to 0} \frac{x \cdot \sin ((x))}{2 - 2 \cos ((x))}$ , is heading to zero.

$\lim_{x \to 0} g (x) = \lim_{x \to 0} h (x) = L$

$\lim_{x \to 0} g (x) =$

$\lim_{x \to 0} 1 - \frac{x^{2}}{6} = 1 - \frac{{(0)}^{2}}{6} = 1$

$\lim_{x \to 0} g (x) = 1$

$\lim_{x \to 0} h (x) =$

$\lim_{x \to 0} 1 = 1$

$\lim_{x \to 0} h (x) = 1$

$\lim_{x \to 0} g (x) = \lim_{x \to 0} h (x) = L$

$\lim_{x \to 0} g (x) = \lim_{x \to 0} h (x) = 1$

Step 4: Apply the conclusion of the Squeeze Theorem to answer the limit question about the equation in the middle, $\lim_{x \to c} f (x) = L$ .

Since we have shown the two requirements of the Squeeze Theorem have been met:

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

2) $\lim_{x \to 0} g (x) = \lim_{x \to 0} h (x) = L$

We can now draw our final conclusion about

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

$\lim_{x \to 0} f (x) = L$

$\lim_{x \to 0} \frac{x \cdot \sin ((x))}{2 - 2 \cos ((x))} = 1$

Final Result:

$\lim_{x \to 0} \frac{x \cdot \sin ((x))}{2 - 2 \cos ((x))} = 1$

Meaning:

The overall limit a $x$ s approaches $0$ of $\frac{x \cdot \sin ((x))}{2 - 2 \cos ((x))}$ is $y = 1$ .

Previous Topic

Back to Lesson

Next Lesson

Unit 1: Limits and Continuity

Example 1: Squeeze Theorem

Leave a Comment Cancel reply