please help my problem . please ....
limit tan[x]/x
x->infinite
Since $\tan x$ is surjective onto the set of reals, given any $k \in \mathbb{R}$, in each interval $\left(\dfrac{n\pi}{2},\dfrac{(n+1)\pi}{2}\right)$ , there exists $a_n$ such that $\tan(a_n) = ka_n$.
$\displaystyle \lim_{n \to \infty} a_n = \infty$
So, since $\displaystyle \lim_{n \to \infty} \dfrac{\tan(a_n)}{a_n} = k$ you know $\displaystyle \lim_{x \to \infty} \dfrac{\tan x}{x}$ does not exist.
Edit: made the argument more general.
The solution I posted is the correct answer to the problem you posted. I even proved it. If you think it is wrong, then you posted the wrong question. Is the correct question supposed to be $\displaystyle \lim_{x \to 0} \dfrac{\tan x}{x}$? Because that is a very different question. You asked the limit as $x \to \infty$, which does not exist.
Even if it is the greatest integer function, the limit would still be undefined. However, if it is the tangent of the fractional part of x over x, then that is different. Using the squeeze theorem, you have $0 \le \dfrac{\tan \langle x \rangle}{x} < \dfrac{\tan 1}{x}$
The limit as $x \to \infty$ of the LHS and RHS of the inequality are both 0, so by the Squeeze Theorem, so is the middle limit.