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

[fire9132]

Homework Statement

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

Homework Equations

The Attempt at a Solution

-1 ≤ cos\frac{1}{x} ≤ 1 \\<br /> 1 - 2^{x} ≤ (2^{x} -1)cos\frac{1}{x} ≤ 2^{x} - 1 \\<br /> 0 ≤ (2^{x} -1)cos\frac{1}{x} ≤ 0 \\<br /> ∴ \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 :)

 

[Mark44]

Homework Statement

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

Homework Equations

The Attempt at a Solution

-1 ≤ cos\frac{1}{x} ≤ 1 \\<br /> 1 - 2^{x} ≤ (2^{x} -1)cos\frac{1}{x} ≤ 2^{x} - 1 \\<br /> 0 ≤ (2^{x} -1)cos\frac{1}{x} ≤ 0 \\<br /> ∴ \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 :)

It's possible that you inadvertently entered something different into wolfram. Your answer is correct.

 

[jbunniii]

-1 ≤ cos\frac{1}{x} ≤ 1 \\<br /> 1 - 2^{x} ≤ (2^{x} -1)cos\frac{1}{x} ≤ 2^{x} - 1 \\<br /> 0 ≤ (2^{x} -1)cos\frac{1}{x} ≤ 0 \\

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.

 

[shortydeb]

1 - 2^{x} ≤ (2^{x} -1)cos\frac{1}{x} ≤ 2^{x} - 1 \\

for x < 0, this statement is false.