# What is the value of $\lim_{x\to 0} \frac{\cos(1/x)}{\cos(1/x)}$?

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

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.

• Yes, this function is equal to the constant function $1$ on $x\neq0$. Their limits are the same. May 16 '19 at 12:17
• 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 $f(x) = 1$. 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.