Limiting value and language confusion I came across following:
We have $f(a)=|a|^b$ we have to compute limiting value of $f(a)$ as $a\rightarrow 0$ but $a\in \mathbb{R}$\ $\{0\}$ . I want to say that when $b=0$, do I say $f(a)=1$ or do I say $f(a)\rightarrow 1?$ 
The problem is $f(a)$ does not depend on the limit value of $a$ in this case but the question asks about limiting value of $f(a)$.
EDIT I can get some idea from below but may be this seems unclear. Clearly, $f(a)$ is a function and $a$ is a variable. The point is you have to classify the limit of $f(a)$ as $a\rightarrow 0$ depending on value of $b$. My point is that in case $b=0$, $f(a)=1$ regardless of limit of a i.e. $$\lim_{a\rightarrow 0}f(a)=\lim_{a\rightarrow -20}f(a)=\lim_{a\rightarrow 100000}f(a)=\lim_{a\rightarrow k}f(a) =1$$ So saying limit of $f(a)$ makes little sense.
Thank You.
 A: What you're being asked to compute is $\lim_{a \to 0} f(a)$.  Here, $a$ is just a dummy variable.  There is no fixed "a" in question.  You rewrite this as $\lim_{x \to 0} |x|^b$ if it makes it easier for you.
A: $a$ is the variable here, and the value of $|a|$ becomes increasingly smaller as $a \to 0$. $b$ is the constant here. The value of $b$ does not vary as $a \to 0$.
You need to give the value as $f(a) \to 0$. You are correct that the question asks for the limit as $\bf{ a \to 0}$, not for the value at $a = 0$, that is, recalling that $a$ is the variable in this problem.
Even if the domain of $a = x \in \;\mathbb{R}\setminus \{0\},\,$ the limit still exists. What matters is the value of the limit as the variable approaches $0$, not the value of the function at $\,0.$

Per question edit: 
Note that $b$ is a constant, in this problem. You don't know that $b$ is $0$. $b$ could be any value, and that would not change the value of $\;\lim_{a \to 0}\,|a|^b\;$. 
But you do know that as $a \to 0$ from the left and from the right, it gets very, very,ver close to $0$, so you can evaluate the limit as $|a|^b \to |0|^b = 1$: which is the limiting value of the function as $a \to 0$. That does not depend on whether it's actually the case that $a = 0$.
