Basic question regarding limits I'm a little confused when it comes to question like this.
Let's say we got this expression: 
$$\lim_{x\to\infty} \tan\left(\frac{1}{x}\right).$$
Am I allowed to say the result of this is  $0$ or do I have to show it? 
(If I have to show it, please show me how to cuz I got no clue).
And more general, in which situations are we allowed to "cut" it and in which we are not? (I'm guessing whenever it comes to fractions surely).
 A: That depends.
If the whole exercise is just "compute $\lim \tan(1/x)$", then yes, you have to do some argument. Something along the lines of "if $x$ goes to infinity, then $1/x$ goes to zero, and by continuity of $\tan$, the whole expression then goes to zero".
But if this limit is just one part of a longer question/calculation, then no. It's trivial enough that anybody with some math background will see it immediately.
A: You need to show that given any small $\epsilon > 0$, there exists a $X(\epsilon)$ such that $|\tan(1/x)| < \epsilon,\; \forall x >  X(\epsilon) $. To do this choose, for e.g., $X(\epsilon) = \dfrac{1+\sec \epsilon}{\epsilon}$. Then 
$$\left|\tan \frac{1}{x}\right| < \left|\tan \frac{\epsilon}{1+\sec \epsilon} \right| = \frac{\sin \left( \frac{\epsilon}{1+\sec \epsilon}\right) }{\cos\left( \frac{\epsilon}{1+\sec \epsilon}\right)} < \frac{\sin \left( \frac{\epsilon}{1+\sec \epsilon}\right) }{\cos\epsilon} < \frac{ \frac{\epsilon}{1+\sec \epsilon} }{\cos\epsilon} = \frac{\epsilon}{1 + \cos \epsilon} < \epsilon$$
A: The answer is that it depends on the situation.  To address the statement in the OP "If I have to show it, please show me how to cuz I got no clue," I thought it might be instructive to present an approach that relies on a standard inequality from elementary geometry.  To that end, we begin with a short primer.


PRIMER:
Recall from elementary geometry the inequality
$$\sin(\theta)\le \theta\tag 1$$
for $\theta\ge 0$.  Squaring both sides of $(1)$, using $\sin^2(\theta)=1-\cos^2(\theta)$, and rearranging, we find that 
$$\cos(\theta)\ge \sqrt{1-\theta^2} \tag 2$$
for $0\le \theta \le 1$.  


Now, using $(1)$ and $(2)$ with $\theta=1/x$ reveals that 
$$\begin{align}
\tan(1/x)&=\frac{\sin(1/x)}{\cos(1/x)}\\\\
&\le \frac{1/x}{\sqrt{1-\frac1{x^2}}}\\\\
&=\frac{1}{\sqrt{x^2-1}}\tag 3
\end{align}$$
for $x>1$.

Hence, for all $\epsilon>0$, we have from $(3)$ that 
$$\begin{align}
\tan(1/x)&\le \frac{1}{\sqrt{x^2-1}}\\\\
&<\epsilon
\end{align}$$
whenever $x>\sqrt{1+\frac{1}{\epsilon^2}}$.  And we are done!
A: Short answer is $0$ ,but if you want  " to show " recall the limit $$\lim _{ x\rightarrow \infty  }{ \frac { 1 }{ x } =0 } $$
A: You can move the limit into the $tan$, because you are not excluding any $x$ from the limit, and you are not creating any limits that don't exist. Once that is done, the problem is trivial.
