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

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.

StubbornAtom
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Aryan Ishan
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  • Yes, this function is equal to the constant function $1$ on $x\neq0$. Their limits are the same. – logarithm May 16 '19 at 12:17
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    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 $f(x) = 1$. – Matti P. May 16 '19 at 12:18

1 Answers1

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$\dfrac{\cos(1/x)}{\cos(1/x)}$ 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).

hmakholm left over Monica
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