Using the squeeze principle to evaluate the following limit

Question

This is my teacher's work. I am stuck on the part where I circled it red. From what I understand, he squared every side, so (-1)^2, the cos^2, and the 1^2. Why did (-1)^2 change to 0?

Answer 1

the $-1$ did not turn into $0$, the inequality $0\leq \cos^2 (3x+1)$ comes from the fact that $x^2\geq0$ if $x$ is a real number.

The statement $0\leq \cos^2(3x+1) \leq 1$ is actually composed of two inequalities. One, $\cos^2(3x+1)\leq 1$, comes from the fact that $\cos(3x+1)$ is between $-1$ and $1$. The other, $\cos^2(3x+1)\geq 0$, comes from the fact that $\cos^2(3x+1)$ is the square of a real number, and squares of all real numbers are non-negative.

Answer 2

Notice that $\cos(a) = 0$ when $a = \frac{\pi}{2} + n\pi$ for $n \in \mathbb{Z}$.

Thus, $\cos^2(a) = 0$. What does this say about the interval in question?

Another way to look at it is that the cosine function can evaluate to any value in $[-1,0] \cup [0,1]$. So when we square it, it becomes $[0,1]$.

What values are between $-1$ and $1$? Well, there's $-0.9$, which squares is $0.81$. There's $-0.5$, which squares is $0.25$. If we keep going we eventually get to $0$, which squared is still $0$. Then we have all the values from $0$ to $1$.

So as you can see, the interval becomes $[0,1]$. No matter how you look at it.

Answer 3

Exercise: Let $a\geqslant0$ and $b\geqslant0$, then the following equivalences hold.

• $a\leqslant x\leqslant b\iff a^2\leqslant x^2\leqslant b^2$
• $-a\leqslant x\leqslant b\iff 0\leqslant x^2\leqslant\max(a,b)^2$

Hint: Plot a graph of the function $x\mapsto x^2$, say on the interval $[-4,3]$.