Squeeze Theorem

[Needhelp123]

Wondering if someone can help with this question:

Use the squeeze theorem to find lim x->0 for Sqrt (x^2 cos^2 1/x)

Thanks!

 

[pka]

Wondering if someone can help with this question:
Use the squeeze theorem to find lim x->0 for Sqrt (x^2 cos^2 1/x)

Here are some facts: \(\displaystyle \sqrt {{x^2}{{\cos }^2}({x^{ - 1}})} = \left| x \right|\left| {\cos ({x^{ - 1}})} \right|\) also \(\displaystyle \left| {\cos ({x^{ - 1}})} \right| \le 1\)

Combine those two facts and let \(\displaystyle x\to 0\). What do you get?

 

[Needhelp123]

Here are some facts: \(\displaystyle \sqrt {{x^2}{{\cos }^2}({x^{ - 1}})} = \left| x \right|\left| {\cos ({x^{ - 1}})} \right|\) also \(\displaystyle \left| {\cos ({x^{ - 1}})} \right| \le 1\)

Combine those two facts and let \(\displaystyle x\to 0\). What do you get?

I understand the part about the absolute values and that cos x^-1 is the equivalent to cos 1/x. I'm guess I'm struggling to understand what cos x^-1 means in relation to cos x. I apologize if it's a very simple answer, but I've been trying to make sense of this all day, and unfortunately nothing seems to be making sense anymore!

 

[pka]

I understand the part about the absolute values and that cos x^-1 is the equivalent to cos 1/x. I'm guess I'm struggling to understand what cos x^-1 means in relation to cos x. I apologize if it's a very simple answer, but I've been trying to make sense of this all day, and unfortunately nothing seems to be making sense anymore!

I really don't what you are saying there. What \(\displaystyle \cos (x)~?\) There is none in the question that I can see.

\(\displaystyle -1\le \cos (x)\le 1\) for all \(\displaystyle x\) so \(\displaystyle - 1 \le \cos \left( {\frac{1}{x}} \right) = \cos ({x^{ - 1}}) \le 1\).

Which give \(\displaystyle \left| {\cos ({x^{ - 1}})} \right| \le 1\) _thus \(\displaystyle \left| {\cos ({x^{ - 1}})} \right|\left| x \right| \le \left| x \right|\)

 

[Needhelp123]

I really don't what you are saying there. What \(\displaystyle \cos (x)~?\) There is none in the question that I can see.

\(\displaystyle -1\le \cos (x)\le 1\) for all \(\displaystyle x\) so \(\displaystyle - 1 \le \cos \left( {\frac{1}{x}} \right) = \cos ({x^{ - 1}}) \le 1\).

Which give \(\displaystyle \left| {\cos ({x^{ - 1}})} \right| \le 1\) _thus \(\displaystyle \left| {\cos ({x^{ - 1}})} \right|\left| x \right| \le \left| x \right|\)

Sorry to be confusing! I was just trying to make sense of what cos x^-1 meant and what it would look like and was drawing a blank! Simple, but it wasn't making sense! Got it, and what you've added helps clarify things. Thank you!