As the title says, I want to show that the limit of
$$\lim_{x\to 0} \frac{\cos(x)}{x}$$
doesn't exist.
Now for that I'd like to show in a formally correct way that
$$\lim_{x\to 0^+} \frac{\cos(x)}{x} = +\infty$$
I'm sure this is right since $\displaystyle\lim_{x\to 0^+} \cos(x) = 1$ and $\displaystyle\lim_{x\to 0^+} x = 0$, but since $\displaystyle\lim_{x\to 0^+} x = 0$ I can't just say:
$$\lim_{x\to 0^+} \frac{\cos(x)}{x} = \frac{1}{\displaystyle\lim_{x\to 0^+} x}$$
Things I tried: expand the fraction and use L'Hospital's rule (This didn't seem to yield results even though I tried quite a few things, should it? If it does, should it always work? It worked for similar problems) and going by the definition of the limit which didn't work for me either.
I'm grateful for any hints for doing this problem and similar problems especially.
$\endgroup$ 22 Answers
$\begingroup$Hint: You can find a region near $x=0$ where $\cos x \gt \frac 12$
$\endgroup$ $\begingroup$If $|x| \in [0,\frac{\pi}{4}]$, then $\cos x \ge \frac{1}{2}$. Then for $n>1$ we have $\frac{1}{n} \in [0,\frac{\pi}{4}]$, hence setting $x_n = \frac{1}{n}$ we have $\frac{\cos x_n}{x_n} \ge \frac{1}{2 x_n} = \frac{n}{2}$.
Similarly, with $x'_n = -\frac{1}{n}$, $\frac{\cos x'_n}{x'_n} \le \frac{1}{2 x'_n} = -\frac{n}{2}$.
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