# Compute $\lim_{x\to0^+}⁡ \sqrt{x\left[\cos^2\left(\frac1{x^3}\right)-3\right]}$ using Squeeze theorem

This is for a first year calc course, and we are using the squeeze theorem to determine the limits of composite functions containing trig functions. I'm familiar with the examples posted in other questions on this website, but haven't seen anything addressing the squeeze theorem as applied to trig functions squared.

My question is how to use the squeeze theorem on the example below:

$$\lim_{x\to0^+}⁡ \sqrt{x\left[\cos^2\left(\frac1{x^3}\right)-3\right]}$$

Normally, I would set the upper and lower bounds of this as being $$-1 ≤ f(x) ≤ 1$$ because that is the range of the cosine function, but I wasn't sure how the $$\cos^2$$ would affect this. Any help would be greatly appreciated!

• Welcome to MSE. Please use MathJax to format both, your questions and your answers :) – mrtaurho Oct 11 '18 at 22:04
• @mrtaurho, I'm not sure the OP intended the square root to extend over the entire expression. They might be asking about the limit from the right of $\sqrt x[\cos^2(1/x^3)-3]$, possibly even with the square brackets denoting the greatest integer function. – Barry Cipra Jan 26 '20 at 6:01

hint

For $$x\ne 0,$$

$$0\le \cos^2(\frac{1}{x^3})\le 1$$

but your function is not defined on the right of $$0$$. you cannot compute $$\lim_{x\to0^+}f(x)$$.

• Thank you! but how would you know it is zero on the left? Do we have to do something algebraically? – jaws0 Oct 11 '18 at 22:05
• @jaws0 The square of a real is forever positive. – hamam_Abdallah Oct 11 '18 at 22:06
• oh, so the lowest the function could ever go is being equal to or greater than 0? thats why the lower boundary is zero, and the upper is still one? – jaws0 Oct 11 '18 at 22:07

Let's define your function as $$f(x) = \sqrt{x\cos^2(x^{-3}) - 3x}$$.

One way to think about it is to consider $$f$$ as a composition of several functions, and just "build up" the necessary inequality by bounding one of the inner functions, and transforming the inequality by each composing function, being sure to flip the signs whenever multiplying by a negative/composing by any decreasing function.

Specifically, for $$x \le 0$$: $$-\infty \le x^{-3} \le \infty \\ -1 \le \cos(x^{-3}) \le 1 \\ 0 \le \cos^2(x^{-3}) \le 1 \\ 0 \ge x\cos^2(x^{-3}) \ge x \\ -3x \ge x\cos^2(x^{-3}) - 3x \ge -2x \\ \sqrt{-3x} \ge \sqrt{x\cos^2(x^{-3}) - 3x} \ge \sqrt{-2x}$$

So you can use the squeeze theorem between the functions $$\sqrt{-3x}$$ and $$\sqrt{-2x}$$ to show that the limit from the left is $$0$$.

Regarding your question about the range of $$\cos^2(x)$$... without worrying about cosine for just a moment, how would the function $$g(x) = x^2$$ affect the real interval $$[-1, 1]$$?

In the same vein as above - think of $$\cos^2$$ as a composition of two functions, $$\cos^2 = h\circ g$$, where $$g : x \mapsto x^2$$, and where $$h : x \mapsto \cos(x)$$.

• The image of the real number line $$\mathbb{R}$$ under $$h$$ is: $$\;h(\mathbb{R}) = [-1, 1]$$.
• The image of $$[-1, 1]$$ under $$g$$ (think of the range along the $$y$$-axis of a parabola from $$(-1, 1)$$ down to $$(0, 0)$$ and back up to $$(1, 1)$$) is: $$\;g([-1, 1]) = g(h(\mathbb{R})) = [0, 1]$$.
• Thus, the image of $$\mathbb{R}$$ under $$h \circ g$$ is: $$\;(h \circ g)(\mathbb{R}) = [0, 1]$$.