# Prove that the limit doesn't exist of the combination of sines and cosines

Prove that $$\lim_{x \to 0} \left[\dfrac{(2 x^2-1) \sin\left(\frac{1}{x}\right)-2 x \cos(\frac{1}{x})}{x^3} \right]$$ doesn't exist.

I am quite sure that this limit can't exist since $\lim_{x \to 0} \dfrac{\sin{1/x}}{x^m}$ doesn't exist but I would like some verification. In general if we are evaluating a limit and one of the terms limit doesn't exist does that mean the entire limit doesn't exist either?

• Generally speaking it IS possible for $\lim_{x\rightarrow a}f(x)$ not to exist but for $\lim_{x\rightarrow a}[f(x)+g(x)]$ to exist. – christina_g Mar 18 '16 at 17:33
• For instance, if $f(x) = -g(x)$... – Clement C. Mar 18 '16 at 17:33
• As $x$ tends to 0, $\sin\frac{1}{x}$ oscillates between 0 and 1. The other terms in the numerator tend to zero, so the numerator is oscillating between values greater than 0.5 and less than -0.5. But the denominator tends to 0. Hence the whole expression is oscillating between ever larger positive and negative values. So the limit cannot exist. – almagest Mar 18 '16 at 17:33

Consider the sequence $(x_n)_n$ defined by $$x_n = \frac{1}{2\pi n+ \frac{\pi}{2}}$$ for $n \geq 1$. Note that $x_n \xrightarrow[n\to\infty]{} 0$.

Then, for your function defined by $f(x) = \frac{(2x^2-1)\sin\frac{1}{x}-2x\cos\frac{1}{x}}{x^3}$ (for $x\neq 0$), one has $$f(x_n) = \frac{2x_n^2-1}{x_n^3} \xrightarrow[n\to\infty]{} -\infty$$ so $f$ cannot converge at $0$.

(Note that by also considering $y_n = \frac{1}{2\pi n + \pi}$, one can also show that $f$ has no limit at all even in $\mathbb{R}\cup\{\pm\infty\}$, i.e. it does not diverge to $-\infty$ either: since $f(y_n)\xrightarrow[n\to\infty]{} +\infty$.)

Hint: Evaluate the function at $1/(2\pi n),$ $n$ an integer.

Take the sequence

$$\left\{\,y_n:=\frac2{(2n-1)\pi}\,\right\} \implies$$

$$f(y_n)=\frac{\left(\frac4{(2n-1)^2\pi^2}-1\right)\overbrace{\sin\frac{2n-1}2\pi }^{=1}-\frac4{(2n-1)^2\pi^2}\overbrace{\cos\frac{2n-1}2\pi}^{=0}}{\frac8{(2n-1)^3\pi^3}}=$$

$$=\color{red}{\pm}\frac{4(2n-1)n\pi-(2n-1)^3\pi^3}{8}\xrightarrow[n\to\infty]{}\text{doesn't exist}$$

and thus the limit cannot exist (observe that $\;\sin\frac{2n-1}2\pi=1\;$ iff $\;n=1\pmod2\;$)

• The last equality is wrong: you forgot the $-1$ term, which dominates. Also, the $\sin$ will not always be $1$, it's going to be $(-1)^{n+1}$ with your choice of $y_n$. – Clement C. Mar 18 '16 at 17:46
• @ClementC. Thank you very much. You're right, yet I think it is easy to mend. Right now. – DonAntonio Mar 18 '16 at 18:05

Best way is to analyze each part of the complicated expression. Let's start with the simplest. Since $\cos(1/x)$ is bounded therefore $2x\cos(1/x)$ tends to $0$ as $x \to 0$. Again $(2x^{2}-1) \to -1$ as $x \to 0$ and $\sin(1/x)$ oscillates between $-1$ and $1$ therefore $(2x^{2}-1)\sin(1/x)$ oscillates between $-1$ and $1$. It follows that the function $$f(x) = (2x^{2} - 1)\sin(1/x) - 2x\cos(1/x)$$ oscillates between $-1$ and $1$. The function $g(x)=1/x^{3}$ tends to $\infty$ or $-\infty$ as $x\to 0^{+}$ or as $x \to 0^{-}$. Therefore the product $f(x)g(x)$ oscillates infinitely as $x\to 0$. Therefore the limit in the question does not exist.