$$\lim_{x\to 0} \frac{\cos\left(\frac{1}{x}\right)}{\cos\left(\frac{1}{x}\right)}$$

I think the answer should be $1$ , I understand that the value of $\cos(1/x)$ would be oscillating quickly as $x$ approaches $0$ . But , wouldn't both the numerator and denominator get cancelled ?

My teacher said that the limit does not exist in this question , I dont understand.

  • $\begingroup$ Yes, this function is equal to the constant function $1$ on $x\neq0$. Their limits are the same. $\endgroup$
    – logarithm
    May 16 '19 at 12:17
  • 1
    $\begingroup$ If $x \neq 0$, $\frac{\cos{\frac{1}{x}}}{\cos{\frac{1}{x}}} = \frac{1}{1} = 1$, and therefore outside of $x=0$ you can replace the whole function by the constant function $f(x) = 1$. $\endgroup$
    – Matti P.
    May 16 '19 at 12:18

$\dfrac{\cos(1/x)}{\cos(1/x)}$ is just a convoluted way of writing the constant function $1$ defined on the domain $$\mathbb R \setminus\{0\} \setminus \Bigl\{\frac1{\pi/2+k\pi} \Bigm| k\in\mathbb Z \Bigr\} $$ Whether this function has a limit for $x\to 0$ depends more on which precise conventions you use for limits, than on the function itself.

In most of higher mathematics we'd have no problem speaking of a limit towards any limit point of the function's domain, and in that case the limit is easily $1$.

On the other hand, many introductory texts try to avoid confusion by insisting that $\lim\limits_{x\to a} f(x)$ is only defined when the domain of $f$ contains an entire punctured neighborhood of $a$ (as a subset of $\mathbb R$). If that convention is used, your limit doesn't exist, and it seems your teacher is assuming that way of thinking.

(Whether this convention actually avoids any confusion is debatable).


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