# How to solve a limit to negative infinity of a function with cos and sin?

Does anyone know how to solve the following limit? I've tried using the squeeze theorem, or rewriting it, but I seem to get nowhere when doing that. Does anyone know how one would go about solving something like this?

$$\lim_{x\to-\infty} \frac{x\cos(x^{2})+3x\sqrt{1-4x}}{\sqrt[4]{x^{6}-x^{5}}+\sin(x^{5})}$$

• Typically all you can know is that $\sin(~)$ is going to return a number somewhere between $-1$ and $1$. You should hope that everything else is going to overshadow that, which is indeed what happens here. Commented Oct 9, 2023 at 19:52
• More precisely: since $x$ is getting large (in absolute value), what is the largest power of $x$ that appears in the numerator? What is the largest power of $x$ that appears in the denominator? Those should be the dominant terms. Commented Oct 9, 2023 at 20:54

A simple approach is to treat $$\cos$$ and $$\sin$$ as if they were just random number generators that return a value in the interval $$[-1, 1]$$.

So, for the numerator of your limand, $$N(x) := x\cos(x^{2})+3x\sqrt{1-4x}$$, you have:

$$x + 3x\sqrt{1-4x} \le N(x) \le -x+3x\sqrt{1-4x}$$

(Since you're taking the limit as $$x$$ goes to negative infinity, we can assume $$x < 0$$, hence having $$x$$ on the left and $$-x$$ on the right.)

And for the denominator, $$D(x) := \sqrt[4]{x^{6}-x^{5}}+\sin(x^{5})$$, we have:

$$\sqrt[4]{x^{6}-x^{5}} - 1 \le D(x) \le \sqrt[4]{x^{6}-x^{5}}+1$$

The left-hand side of the inequality is zero if $$x \approx -0.8812714616335696$$. But if $$x$$ is below that value, then $$D(x)$$ is positive. Thus, as $$x \to -\infty$$, we have:

$$\frac{1}{\sqrt[4]{x^{6}-x^{5}} - 1} \ge \frac{1}{D(x)} \ge \frac{1}{\sqrt[4]{x^{6}-x^{5}}+1}$$

Combining this with the previous inequality for $$N(x)$$, we get:

$$\frac{x + 3x\sqrt{1-4x}}{\sqrt[4]{x^{6}-x^{5}}+1} \le \frac{N(x)}{D(x)} \le \frac{-x + 3x\sqrt{1-4x}}{\sqrt[4]{x^{6}-x^{5}}-1}$$

You should be able to show that the bounding functions $$L(x) := \frac{x + 3x\sqrt{1-4x}}{\sqrt[4]{x^{6}-x^{5}}+1}$$ and $$U(x) := \frac{-x + 3x\sqrt{1-4x}}{\sqrt[4]{x^{6}-x^{5}}-1}$$ have the same limit, and then the Squeeze Theorem gives you the limit of the original expression.

• This was what I originally had done, but I wasn't able to properly work it out. I eventually ended up that the limit goes to negative infinity, which I don't think is quite true, but I cannot determine where I went wrong. What steps do you take from here on out? Commented Oct 9, 2023 at 21:47
• @RonanFinn: You can use L'Hospital's Rule. Or note that both N(x) and D(x) are asymptotically $O(x^{3/2})$...
– Dan
Commented Oct 9, 2023 at 21:53
• @Dan I like how you formalized what I hand-waved on the oscillating terms.. Commented Oct 9, 2023 at 22:32

Following the above comment, without trying to give a formal proof with $$L$$ and $$\epsilon$$, you can break this down and just focus on powers of $$x$$ in the top and bottom (not that different than a simpler problem of a ratio of polynomials as $$x\rightarrow+\infty$$).

Divide by $$x$$.

Now the numerator is:

$$cos(x^2)+3\sqrt{1-4x}$$

And the denominator is:

$$\frac1x\space\sqrt[4]{x^6-x^5} + \frac{sin(x^5)}x$$

The $$cos$$ term is bounded, and the term with $$sin$$ approaches $$0$$. So we can focus on the ratio of the others. [This is handled with more detail in the above subsequent answer.]

As $$x\rightarrow-\infty$$:

The remaining top term approaches $$6\sqrt{-x}=6\sqrt{|x|}$$.

And the remaining bottom term approaches $$\frac1x |x|^{3/2}=\frac1x|x|\sqrt{|x|}=-\sqrt{|x|}$$.

Therefore the original expression approaches $$-6$$ as $$x\rightarrow-\infty$$.

• The third- and second-to-last-lines are vague enough to be misleading. One function doesn't "go to" another function. Commented Oct 9, 2023 at 20:55
• @GregMartin this was supposed to be a bit heuristic in line with simpler problems where you look at highest powers; is "approaches as x goes to neg. infinity" better or are you saying this is not really valid reasoning? (appreciate the feedback BTW). Commented Oct 9, 2023 at 20:59
• I think it's better to divide both top and bottom by $|x|^{3/2}$ rather than $x$. Then every individual term tends to a finite limit as $x\to-\infty$, and the arithmetic limit laws apply rigorously. Commented Oct 9, 2023 at 22:39
• Fair enough, and in retrospect yes. Started with $x$ since it was common in the numerator and helped do away with the trig functions. Commented Oct 9, 2023 at 22:44
• Well, I would propose that "the sin and cos terms are bounded so can effectively be ignored" is also vague enough to be misleading. Commented Oct 10, 2023 at 6:04