Solve $\lim_{x\to0} \tan^{-1}\frac{1}{x^2}$ Solve $\lim\limits_{x\to0} \tan^{-1}\dfrac{1}{x^2}$
We know that to find limit at $x=0$, function must be defined in neighborhood of $x=0$
Let's see if function is defined in left neighborhood of $x=0$
$l=0-\delta$ will be the left neighborhood of $x=0$ where $\delta$ is very very$\cdots\cdots$ small positive number.
$$\tan^{-1}\dfrac{1}{(0-\delta)^2}$$
$$\tan^{-1}(+\infty)$$
As we know $\tan^{-1}$ is not defined at $+\infty$, so it means at left neighborhood function is not defined.
Let's see if function is defined in right neighborhood of $x=0$
$r=0+\delta$ will be the right neighborhood of $x=0$.
$$\tan^{-1}\dfrac{1}{(0+\delta)^2}$$
$$\tan^{-1}(+\infty)$$
As we know $\tan^{-1}$ is not defined at +$\infty$, so it means at right neighborhood also function is not defined.
But actual answer is $\dfrac{\pi}{2}$ which is not difficult to find why but it breaks the basic principle of limits right? What am I missing here?
 A: We have that 
$$f(x) =\dfrac{1}{x^2} \to \infty$$
and then
$$\lim_{x\to 0}\arctan (f(x)) =\lim_{f(x)\to \infty}\arctan (f(x)) = \frac \pi 2$$
A: You had better get a clear idea of the term neighborhood which seems to be primary source of confusion (as indicated by your comments).
A neighborhood of a point $c$ is an open interval containing $c$. Thus $(1,3), (1,2.001)$ are both neighborhoods of point $2$. In your comments you have developed a notation for right neighborhood and its definition should be $$c^{+} =I_c\cap\{x\mid x\in\mathbb {R}, x>c\} $$ where $I_c$ is any neighborhood of $c$. Thus a right neighborhood of $c$ is a neighborhood with all points less than or equal to $c$ removed from it. By the very nature of this definition your notation $c^{+} $ does not represent any one specific set but rather can be used to represent any right neighborhood.
It should be also be observed that $c^{+} $ is always of the form $(c, d) $ or $(c, \infty) $. 
The key point is that $c\notin c^{+} $ and thus if $x\in 0^{+}$ then $x\neq 0$ and hence $1/x^2$ is well defined and consequently $\tan^{-1}(1/x^2)$ is well defined. 

Another typical problem is the use of the term "very small number" (repeat the word very as many times as you wish to increase its intensity or effect). The phrase is meaningless without any additional context. A number is smaller compared to another number.
In typical daily life real world scnearios we know the context and can use the phrase "very small" accordingly.
In the context of real numbers there is no number which is very small. There is always a smaller one. This is an intuitive fact but somehow people learning calculus don't seem to appreciate it. And often one desperately desires the existence of a very very small positive number. This line of thinking has to be ditched seriously. 
A: $\mathbf{\text{Hint:}}$
Since $1/x^2\gt 0$ hence $$\arctan\left(\frac{1}{x^2}\right)=\frac{\pi}{2}-\arctan(x^2)$$
A: HINT: $$\lim_{x\to 0}\tan^{-1}\left(\frac{1}{x^2}\right)=
\lim_{x\to 0}\cot^{-1}\left(x^2\right)$$
Where, $\cot^{-1}\theta\in(0, \pi)\quad \forall \ \theta\in(-\infty, \infty)$
