# Squeeze Theorem/Limits Question. Answers don't match up.

1. Jun 17, 2013

### fire9132

1. The problem statement, all variables and given/known data

Evaluate using the Squeeze Theorem.
$\lim_{x\rightarrow0} (2^{x} - 1) cos\frac{1}{x}$

2. Relevant equations

3. The attempt at a solution

$-1 ≤ cos\frac{1}{x} ≤ 1 \\ 1 - 2^{x} ≤ (2^{x} -1)cos\frac{1}{x} ≤ 2^{x} - 1 \\ 0 ≤ (2^{x} -1)cos\frac{1}{x} ≤ 0 \\ ∴ \lim_{x\rightarrow0} (2^{x} - 1) cos\frac{1}{x} = 0$

That's the solution I got and the solution that's in my textbook. But when I checked on wolframalpha, it says the limit does not exist.

Did I do something wrong and is my textbook wrong? or is it something conceptually that I don't understand about this? Is wolframalpha wrong?
Thanks :)

2. Jun 17, 2013

### Staff: Mentor

3. Jun 18, 2013

### jbunniii

You are correct that the limit is zero, but your reasoning has some flaws. The last line is clearly not correct, for it is equivalent to
$$(2^x - 1)\cos\left(\frac{1}{x}\right) = 0$$
which is not true for all $x$. The previous line is correct, but only if $2^x - 1 \geq 0$.

It's somewhat easier to work with absolute values:
$$\left|\cos\left(\frac{1}{x}\right)\right| \leq 1$$
Now multiply both sides by $|2^x - 1|$ and see what you can conclude.

4. Jun 19, 2013

### shortydeb

for x < 0, this statement is false.