0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – christina_g Mar 18 '16 at 17:33
  • $\begingroup$ For instance, if $f(x) = -g(x)$... $\endgroup$ – Clement C. Mar 18 '16 at 17:33
  • $\begingroup$ 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. $\endgroup$ – almagest Mar 18 '16 at 17:33
1
$\begingroup$

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 $f(y_n)\xrightarrow[n\to\infty]{} +\infty$.)

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $\;n=1\pmod2\;$)

$\endgroup$
  • $\begingroup$ 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 $y_n$. $\endgroup$ – Clement C. Mar 18 '16 at 17:46
  • $\begingroup$ @ClementC. Thank you very much. You're right, yet I think it is easy to mend. Right now. $\endgroup$ – DonAntonio Mar 18 '16 at 18:05
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.