How to determine the limits of $\cos(x/3) / \cos(x)$ over the domain $x \in [0, \pi/2]$?

Question

I'm trying to determine the range of $y = \frac{\cos(x / 3)}{\cos(x)}$ over the domain $x \in [0, \pi / 2]$.

This is my attempt:

$$ \lim_{x \rightarrow 0} \frac{\cos(x / 3)}{\cos(x)} = \frac{\cos(0 / 3)}{\cos(0)} = 1 $$

and

$$ \lim_{x \rightarrow \pi / 2} \frac{\cos(x / 3)}{\cos(x)} = \lim_{x \rightarrow \pi / 2} \frac{\sin(x / 3)}{\sin(x)}{1 \over 3} = \frac{\sin(\pi / 6)}{\sin(\pi / 2)}{1 \over 3} = 1 / 6 $$

(where L'Hopital's rule is used in the second equation)

So this gives $1/6 < y < 1$.

However, the graph of $y$ shows that $y$ ranges from $1$ to $\infty$ over the domain $x \in [0, \pi/2]$.

Where have I gone wrong in my attempt?

Answer 1

You can't use De L'Hopital theorem because the limit should be in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

By direct evaluation we get

$$ \lim_{x \to \tfrac{\pi}{2}^-} \frac{\cos{(x/3)}}{\cos x} =\frac{\cos{(\pi/6)}}{0^+}=+\infty $$

Answer 2

What if $\cos\dfrac x3=0?$

otherwise,

$$\dfrac{\cos x}{\cos\dfrac x3}=\dfrac1{4\cos^2\dfrac x2-3}$$

$$\text{As }1\ge 4\cos^2\dfrac x2-3\ge-3,$$

For $1\ge 4\cos^2\dfrac x2-3\ge0,$ $$\dfrac1{4\cos^2\dfrac x2-3}\ge1$$

For $0>4\cos^2\dfrac x2-3\ge-3,$ $$\dfrac1{4\cos^2\dfrac x2-3}\le-\dfrac13$$