Which of the following is/are correct ?
$(1)\lim_{x\to 0}\big[\frac x{sinx}\big]=1$
$(2)\lim_{x\to 0}\big[\frac x{cosx}\big]=0$
$(3)\lim_{x\to 0}\big[\frac {sinx}x\big]=0$
$(4)\lim_{x\to 0}\big[\frac {cosx}x\big]=1$
My attempt:-
We have $\lim_{x\to 0}\frac{sinx}x=\lim_{x\to 0}\frac x{sinx}=1$
So for a given $\epsilon \gt 0, \exists \delta \gt 0$ such that
$1-\epsilon \lt \frac{sinx}x,\frac x{sinx}\lt 1+\epsilon, \forall x\in (-\delta,\delta)$ (Actually there are two $\delta$'s , I am selecting the minimum out of them.
Now if $ \sin x \lt$ or $\gt x$ according as $x\gt $ or $\lt 0$.
So for $x\in (0,\delta)$,
$1-\epsilon \lt \frac{\sin x}x \lt 1$
Thus $\lim_{x\to 0+}\big[\frac{\sin x}x \big]=0$
For $x\in (-\delta,0)$
$\sin x\gt x \Rightarrow \frac{\sin x}x\lt 1$ (since $x\lt 0$)
By similar reasoning as above
$\lim_{x\to 0-}\big[\frac{\sin x}x \big]=0$
Hence , $\lim_{x\to 0}\big[\frac{\sin x}x\big]=0$
In the case of $\frac x{\sin x}$ , we would have
$1\lt \frac x{\sin x}\lt 1+\epsilon$, for $x\in (-\delta,\delta)$
Thus $\lim_{x\to 0} \big[\frac x{\sin x}\big]=1$
So $(1)$ and $(3)$ are true
Well for $(4)$ ...
$\cos x\ge 1-\frac{x^2}2,\forall x \in \mathbb{R}$
So for $x\gt 0$, we have
$\frac{\cos x}x\ge \frac 1x-\frac x2\to \infty$ as $x\to 0$
So the limit in $(4)$ does not exist.
For $(2)$,
Now , $\lim_{x\to 0}\frac x{\cos x}=0$
So for arbitary $\epsilon \gt 0, \exists \delta \gt 0$ such that
$-\epsilon \lt \frac x{\cos x} \lt \epsilon$
Now for $x\in (0, \delta), 0\lt \frac x{\cos x}\lt \epsilon $ (since $\cos x\gt 0$ in close neighbourhood of zero) giving
$\lim_{x\to 0+} \big[\frac x{\cos x}\big]=0$
Similarly, $\lim_{x\to 0-} \big[\frac x{\cos x}\big]=0$
Thus $\lim_{x\to 0} \big[\frac x{\cos x}\big]=0$
So $(2)$ is true.
I request you to go through my solution and check if there is any technical difficulty.
I would like any alternative ideas. Thanks for your time.