I was told to evaluate the following limit:
$$\lim_{x \to 0^+} x^{x^2}$$
I know that the limit at 0 is equal to one since $0^0 = 1$, but I don't know what the correct way to do a RHL/LHL is. Do I just sub in a value > 0?
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Whether or not you define $0^0=1$, this has no relevance for the problem at hand, because you are not computing $0^0$, but a limit.
Note that the power $a^x$ (where $x$ is allowed to take any real value) can only sensibly be defined for $a>0$, so the equality $$ a^x=\exp(x\log a) $$ (natural logarithm and standard exponential function) holds.
One could also define $0^x=0$ (for $x>0$), but it would only be marginally useful. And it turns out that the two variable function $$ f(x,y)=x^y $$ defined for $x>0$ and any $y$ has no limit for $(x,y)\to(0,0)$.
This explained, you should always treat limits of the form $$ \lim_{x\to c}f(x)^{g(x)} $$ (two-sided or one-sided) with the following strategy:
compute $\lim_{x\to c}g(x)\log f(x)$
if the limit in 1 exists and is finite, say $l$, then $\lim_{x\to c}f(x)^{g(x)}=e^l$
if the limit in 1 exists and is $\infty$, then $\lim_{x\to c}f(x)^{g(x)}=\infty$
if the limit in 1 exists and is $-\infty$, then $\lim_{x\to c}f(x)^{g(x)}=0$
if the limit in 1 does not exist, then neither the original limit exists.
In this case, $$ \lim_{x\to0^+}x^2\log x=0 $$ which is a basic limit, so indeed $$ \lim_{x\to0^+}x^{x^2}=e^0=1 $$
egreg
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3I wish people really take note of your opening sentence. I could see a lot of fuss on defining $0^{0}$ in comments to other answer which is totally irrelevant here. +1 – Paramanand Singh Apr 20 '17 at 09:09
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@ParamanandSingh: Jaideep's and egrep's solutions are correct. But we want to dispute a claim in a comment that $(=0)^{=0}$ is well defined. I think we all have a duty to find out the truth for the benefit of everyone. – Hoc Ngo Apr 20 '17 at 09:24
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9@HocNgo There's no inherent truth in the matter, it's a question of conventions. Either one defines $0^0$ or one doesn't. The convention that I follow is that $0^0 = 1$. Always. Except when it isn't. – Daniel Fischer Apr 20 '17 at 09:37
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4@TheGreatDuck But that is an entirely different matter. Whether a function is defined at a point is a different question than whether the limit of the function at that point exists. – Daniel Fischer Apr 20 '17 at 15:26
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1@DanielFischer The idea that functions are necessarily continuous is hard to eradicate – egreg Apr 20 '17 at 16:41
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Let $$A=\lim_{x \to 0^+} x^{x^2}$$
Now take logarithm both the sides.
$$\ln A= \lim_{x \to 0^+} x^2 \ln x$$
Now you can use L'Hopitals rule :
$$\ln A = \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x^2}}$$
$$\ln A= \lim_{x \to 0^+} \frac{\frac{1}{x}}{\frac{-2}{x^3}}=\lim_{x \to 0^+} -\frac{x^2}{2} =0$$
$$\ln A =0 \implies A=1$$
Jaideep Khare
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@HenningMakholm (=0)^{(=0)} Isn't defined! Your comment seems to say that it is defined. – Jaideep Khare Apr 20 '17 at 08:15
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@HenningMakholm Is it okay now, since it has nothing to do, I have removed now. – Jaideep Khare Apr 20 '17 at 08:17
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@HenningMakholm, can you point to a reference where $0^0$ is defined? Note that if $n$ is a positive integer, $0^n$ is well defined and is equal to $0$. But $0^0$ can be written as $0^{n-n}$, or $0^n/0^n$, and division by zero is always undefined. – Hoc Ngo Apr 20 '17 at 08:17
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1@HocNgo: $a^0=1$ is part of the definition of exponentiation for every $a$, in every unital ring. Your reasoning would lead to denying that $0^1$ is defined because $0^2/0^1$ divides by zero. – hmakholm left over Monica Apr 20 '17 at 08:19
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3@HocNgo: On the other hand, $a^{x-y}=a^{x}/a^y$ is not a definition, but a property that exponentiation has under certain conditions. Among those conditions is that $a^x$ and $a^y$ are both defined and that one can divide by $a^y$ – hmakholm left over Monica Apr 20 '17 at 08:21
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@HenningMakholm, I agree that my reasoning is flawed when the base is 0. But I still need to see a reference to $0^0$ being a defined value. BTW, how do you apply the L'Hopital's rule in your last expression? I know the limit is zero but the RHS is of the form $0\cdot\infty.$ – Hoc Ngo Apr 20 '17 at 08:24
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@HocNgo By converting it into a fraction of $\infty / \infty $ we can use L'Hopitals rule. See my edit. – Jaideep Khare Apr 20 '17 at 08:28
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1@HocNgo: The fundamental definition of $x^y$ is for $y\in\mathbb N_0$ by the recursion equations $$ x^0 =1 \qquad\qquad x^{y+1} = x^y\cdot x $$ This notion is later extended to larger domains for $y$ by definitions that involve limits, but this extension does not make the values fixed by the original definition go away. And the original definition has nothing do do with limits. – hmakholm left over Monica Apr 20 '17 at 08:28
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@HenningMakholm we're talking about $x^y$ where $x,y\in\Bbb R$ now though, so in that case I guess we agree that we shouldn't assume $0^0=1$. Exponentiation is defined differently when the exponent is a real number rather than a non-negative integer – Apr 20 '17 at 08:39
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@Jaideep, thanks for your edit. I asked this question because what I learned from 45 years ago was that L'Hopital's rule applies to the $0/0$ form only and we students at the time were not allowed to apply to the $\infty/\infty$ form. We had to resort to a change variable to $t=1/x$. So I now learn something new today. – Hoc Ngo Apr 20 '17 at 08:41
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@vrugtehagel: If you don't agree that $0$ is a non-negative integer, then I cannot help you here. – hmakholm left over Monica Apr 20 '17 at 08:43
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I didn't say I didn't agree to that. Of course, $0$ is a non-negative integer. I'm saying you cannot use that definition for real numbers because it's being defined differently. Sure, in most cases, those definitions coincide, but I don't think this is the case where $x=y=0$. – Apr 20 '17 at 08:45
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@vrugtehagel: If you agree that $0$ is a non-negative integer, and that $0^0$ is something raised to $0$, then you must also agree that $0^0$ is something raised to a non-negative integer. – hmakholm left over Monica Apr 20 '17 at 08:47
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@HenningMakholm, I remember learning in the algebra class that $x^0 = 1$ is a convention for all numbers $x$ because it makes the quotient rule consistent. Of course, algebra was taught before calculus and we never considered $0^0$. – Hoc Ngo Apr 20 '17 at 08:58
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@HenningMakholm, Note $x^y$ is first defined for integer $y$ and is later extended by analytic continuation of $y$ to real or complex $y$. If $0^0$ is in deed well defined and has a finite limit, the limiting form $(\to 0)^{(\to 0)}$ must then be well defined as well, simply based on analytic continuation. So if there are examples where $(\to 0)^{(\to 0)}$ is not 1, then I think $0^0$ is not well defined. – Hoc Ngo Apr 20 '17 at 09:00
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Based on wiki page https://en.wikipedia.org/wiki/Indeterminate_form, $0^0$ is one of the indeterminate forms. So $(exact, 0)^{(exact, 0)}$ is indeed undefined. – Hoc Ngo Apr 20 '17 at 09:10
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2@HocNgo: "Indeterminate form" is not the same as "undefined". Indeterminate form means that the limit of $f(g(x),h(x))$ cannot be computed by knowing the limits of $g(x)$ and $h(x)$ separately; it has nothing at all to do with whether the value of $f$ itself is defined at a particular point. – hmakholm left over Monica Apr 20 '17 at 09:21
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@HenningMakholm: If $(=0)^{(= 0)}$ were 1, then the function $f(x,y)=x^y$ could be analytically continued to $(0,0)$ easily, resulting in $f$ being continuous at $(0,0)$. For any continuous function $g(x)$, the limit at $x_0$ is equal to $g(x_0).$ So, in the case of $f(x) = x^{x^2}$, we simply invoke the continuity property to conclude $\lim_{x\to 0} f(x) = 1$ without going through the manipulations shown in Jaideep's and egreg's solutions. – Hoc Ngo Apr 20 '17 at 09:44
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2@HocNgo: The function $(x,y)\mapsto x^y$ is not continuous at $(0,0)$, which invalidates every argument that assumes it is. And the extension of $x^y$ from integer $y$ to real $y$ is not by analytic continuation; you can't even use analytic continuation for a function that is not already defined on an open set (or at least a set with an accumulation point). – hmakholm left over Monica Apr 20 '17 at 09:58
;-)– egreg Apr 20 '17 at 10:36