Find $$\lim_{x\rightarrow0} x\tan\frac1x$$
Now I tried to find the form of the limit ($0/0$ or $0\cdot \infty$ or $\infty/\infty$), but as $x\rightarrow 0$, $\tan(1/x)$ tends to $\tan \infty$, and since $\tan x$ is unbounded unlike $\sin x$ or $\cos x$, no particular value or range can be assumed for $\tan(1/x)$.
Then I tried to find LHL and RHL.
Let $\lim_{x\rightarrow0^+} x\tan{(1/x)}=L$.
Then $\lim_{x\rightarrow0^-} x\tan{(1/x)}=-L$, since $x$ is approaching from the negative side, the input $1/x$ of $\tan$ is the negative of the input in RHL, and $\tan (-x)=-\tan x$
Now if the limit exists, then $LHL=RHL$, thus $L=0$.
Thus I got that if the limit exists, then it must be equal to $0$. But this doesn't confirm that the limit exists (and it doesn't).
Please help me in proving that the limit doesn't exist, and also please point out the mistakes (if any) in the argument I presented above (sorry for I might be weak in limits and the basics of it)
EDIT: As pointed out by Shubham in the comments, I forgot to take the sign of $x$ too in the $LHL$, thus rendering the argument which proved $L=0$ moot.
THANK YOU
Best Answer
Let $f(x)=x\tan(1/x)$ when $x\neq 0.$
If $a_n=\frac{1}{2\pi n}$ then $f(a_n)=0$ for all $n.$ Thus $$\lim_{n\to\infty} f(a_n)=0.$$
On the other hand, let $$b_n=\frac1{2\pi n+\arctan(n)}$$
Then $$\frac{n}{(2n+1)\pi}<f(b_n)<\frac{n}{2n\pi}$$ so by the squeeze theorem, $$\lim_{n\to\infty} f(b_n)=\frac{1}{2\pi}$$
Finally, let $$c_n=\frac1{2\pi n+\arctan(n^2)}$$ and show $f(c_n)\to +\infty.$
But $a_n,b_n,c_n$ all converge to $0,$ so $$\lim_{x\to 0} f(x)$$ cannot exist.
(We only needed any two of these three sequences to disprove convergence.)