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.