# 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.

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Treat $a$ as a variable. –  hjpotter92 Jan 24 '13 at 13:38

$a$ 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 $a = 0$.

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does not limit actually depend on $b$? when $b=0$, it's $1$. when $b$ is sufficiently large i.e. $b\rightarrow\infty$, $\lim=0$ as $a\rightarrow 0$. What happens when both $b\rightarrow 0$ and $a\rightarrow 0$ ? I am asked to classify based on value of $b$. How does that make b totally a constant? –  007resu Jan 24 '13 at 14:49
Ok, Is this correct? $\lim_{a\rightarrow 0}|a|^{\infty}=0$ and $\lim_{a\rightarrow 0}|a|^{0}=1$. Which one is wrong? –  007resu Jan 24 '13 at 14:59
The first one is wrong, since $b \not\to \infty$ (b is a constant, so it cannot be infinity - $\infty$ is not a constant) and since regardless of $b$, $\lim_{a\to 0} |a|^b \to |0|^b \to 1$ –  amWhy Jan 24 '13 at 15:08
Thank You. There was my mistake. –  007resu Jan 24 '13 at 15:09
It turns out here that the value of $b$ doesn't change the value of the limit. –  amWhy Jan 24 '13 at 15:10

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.

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