Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Treat $a$ as a variable. – hjpotter92 Jan 24 '13 at 13:38
up vote 1 down vote accepted

$a$ 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 $a = 0$.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.