They told me that $0^0 = 1$ , but why?

Remember the fact that $\dfrac{1}{\log_bk}=\log_kb.$ Simplifying the expression, $x^\frac{1}{\log x}=x^{\log_xe}=\boxed{e}.$

All that is left to see is that the limit does in fact exist as $x$ approaches $0$ from the right. Observing small values of $x$ such as 0.0001, we see that $0.0001^{\frac{1}{\log0.0001}}$ indeed exist.

@Calvin Lin I believe this is what you were looking for?

Assuming that the limit exists, let,

$L = \lim_{x\to 0} x^{\frac{1}{\log x}}$

Taking $\displaystyle\text{log}$ on both sides,

$\log L = \log\left(\lim_{x\to 0} x^{\frac{1}{\log x}}\right)$

Taking the $\displaystyle \text{log}$ inside the limit,

$\log L = \lim_{x\to 0} \log x^{\frac{1}{\log x}}$

Using the fact that $\displaystyle \log a^b = b\log a$ , we get,

$\log L = \lim_{x\to 0} \frac{1}{\log x} \times \log x$

$\log L = 1$

$L = e$

$\Rightarrow \lim_{x\to 0} x^{\frac{1}{\log x}} = \boxed{e}$

There is a very easy solution, that doesn't require a deep understanding of limits.

Hint: Simplify the expression $x ^ \frac{ 1}{ \log x }$ BEFORE considering limits.

Hint: If you don't know what to do, draw a graph.

had some inspiration from a previous question:

$\lim_{ x \rightarrow 0^+ } x ^ \frac {1} {\ln x}$

let $x = e^u \longrightarrow u = \ln x$

now redefine the limit in terms of our substitution:

$\lim_{ u \rightarrow -\infty } (e^u) ^ \frac {1} {u}$ = $\lim_{ u \rightarrow -\infty } e ^ \frac {u} {u}$ = $\lim_{ u \rightarrow -\infty } e ^ 1$ = $e$

qed

$\lim _{ x\rightarrow { 0 }^{ + } }{ { x }^{ \frac { 1 }{ \ln { x } } } } =\lim _{ x\rightarrow { 0 }^{ + } }{ { e }^{ \ln { \left( { x }^{ \frac { 1 }{ \ln { x } } } \right) } } } =\lim _{ x\rightarrow { 0 }^{ + } }{ { e }^{ \frac { 1 }{ \ln { x } } \ln { x } } } =e$

$x ^ \frac{1}{\ln x}=L$ $\text{So...}$ $(x^ \frac{1}{\ln x})^{\ln x} = L^{\ln x}$ $x=L^{\ln x}$ $\text{Take the } {\ln} \text{ of both sides...}$ ${\ln x} = \ln{(L^{\ln x})}$ $\text{Using } {\log} \text{ laws...}$ ${\ln x} = {\ln x}*{\ln L}$ ${\ln x} - {\ln x}*{\ln L} = 0$ $\text{Factor out a } {\ln x}...$ ${\ln x}(1-{\ln L})=0$ $\text{So either } {\ln x = 0} \text{, or } {\ln L=1},$ $\text{ which means either } {x=1} \text{, or } {L = e.}$ $\text{Since } {x \rightarrow 0^+} \text{, then }{x \neq 1}\text{. Therefore, } {L = e. \boxed{ }}$

I understand I am just level 2 in calculus as per the grading of this site; however, I am of the strong conviction that the answer to this problem is not e . First of all we are being tricked by the author of the problem to believe that since: $\lim_{x \rightarrow 0} x = 0$ and that $lim_{x \rightarrow 0} \frac{1}{lnx} = 0$ therefore the limit of: $x^{\frac{1}{lnx}}$ as $x \rightarrow 0$ = $0^0$ . If this was the case, then there are more than one way to show that $0^0$ is the same as the limit of different other functions, which will be wrong and misleading; an example of such functions is: $x^x$ . According to the author of the problem under question, since: $\lim_{x \rightarrow 0} x = 0$ therefore it means the limit of $x^x$ = $0^0$ . By using any of the many methods in the comments under this question one may be able to find out that the limit as $x \rightarrow 0$ of $x^x$ is equal to $e^0 = 1$ . How?

• $L = \lim_{x \rightarrow 0} x^{x}$

• $\Rightarrow ln L = ln[ \lim_{x \rightarrow 0} x^{x} ]$

using the fact that ln $a^{b}$ = b ln a and also that ln [ lim ] = lim [ ln ] - $\Rightarrow ln L = \lim_{x \rightarrow 0}$ x ln x

$x ln x$ is the same as: $\frac{ln x}{\frac{1}{x}}$ direct substitution of $x=0$ in the limit will yield $\frac{0}{0}$ which means we can use L'Hopital's Rule to easily evaluate the limit. This gives us:

• $ln L = \lim_{x \rightarrow 0} (- x) = 0$

• $\Rightarrow L = e^{0} = 1$

CONCLUSION There are many functions which limits when evaluated are some real values or integers but we can make their limit as x approaches zero appear like $0^{0}$ by wrongly taking their limits in desired parts. There are two examples we have just seen: $x^{x}$ and $x^{\frac{1}{ln x}}$ . Another example could be : ${\frac{1}{ln x}}^{x}$ . All these functions have distinct or perhaps coincidentally the same limits as x approaches zero from the right; it is therefore misleading to conclude that since we can in parts make the limits resemble $0^{0}$ then it means that $0^{0}$ = these different limits.

$0^{0}$ IS NOT EQUAL TO e !!!

just take a subsequence of points $x_n = e^{-n}$ ,as $n \rightarrow \infty$ , $x_n \rightarrow 0$ , so $x^{1/ln(x)}$ becomes $(e^{-n})^{1/ln(e^{-n})}$ = $( e^{-n})^{-1/n}$ , which is $e^{-1*-1} = e^{1} = e$

Make a guess

The function is a constant, which was not obvious to me. Using the inverse functions of natural log and exponential you have

x^ $\frac{1}{ln(x)}$ = e^[ ln( x^ $\frac{1}{ln(x)}$ ) ] = e^[ $\frac{1}{ln(x)}$ *ln(x) ] = e^1

x^ $\frac{1}{ln(x)}$ = e, for x>0

So the right-hand limit is e.

x^y can be written as exp(y log x ).Hence giving us the answer e..

Let $\log x=z$ then $x=e^z; x\rightarrow 0^+\implies z\rightarrow -\infty.$ The limit becomes $\displaystyle\lim_{z\to -\infty}(e^z)^{\frac{1}{z}}=\displaystyle\lim_{z\to -\infty}e=e\quad\square$

x^(1/lnx) is identically equal to e.

lim(e) as x--->o+ = e

sorry, I now see the same solution below

Let $y=\ln{x}$ ; then $x=e^y$ .

Hence $\lim_{x \to 0^+}x^{\frac{1}{\ln{x}}}=\lim_{e^y \to 0^+}{e^y}^{\frac{1}{y}}=\lim_{e^y \to 0^+}e$ which doesn't depend on $y$ .

In other words, $x^{\frac{1}{\ln{x}}}$ is a constant function which is always equal to $e$ .

First we need to assume $ln(x)$ = $y$ and then we can say that $x=e^y$ and and then the expression becomes $e^[y^1/y]$ which is e^y * 1/y which is e

This one was a fairly easy question Lets take y=lim x tends to 0+ (x^(1/lnx) now take ln on both sides we get lny=(1/lnx×lnx) so ln(y) will tend to 1 hence y will tend to e^1 i.e. e.............

NOTE:

$x^{\frac{1}{\ln \left(x\right)}}=e^{\frac{\ln \left(x\right)}{\ln \left(x\right)}}$

i think it's the same as 1^infinity 0^0 is defined to be 1, however, it's different from lim (f(n)^g(n)) , n -> infinity where f(n) & g(n) approaches 0 i think the definition comes from polynomials like P(X) = X^2 + 3 it's actually P(X) = X^2 + 3X^0 and P(0) = 3 = 3 0^0 hence the definition! i also think that 1^infinity is 1 and it's also different from some limit but defining such thing doesn't make sense as infinity is not a number

Can we use the following logic? Let y=limx->0 x^{ $\frac{1}{log x}$ } taking log on both sides log y= lim x->0 $\frac{log x}{log x}$ => log y=lim x->0 (1) => y=e^{1}

Let y=1/logx, then x=e^(1/y),
As x->0+, y->0-,
Now x^(1/logx)=e,
lim x^(1/logx)[when x->0+] =lim e [when y->0-] =e.

Well, this is $e$ , since $\log x^{1/\log x}=\frac{1}{\log x}\log x=1$ . for all $x>0$ . But that does not mean in any way that $\lim\limits_{(x,y)\to 0} x^y=0^0$ . $0^0$ can be (And in fact is) defined as an algebraic result. If we define $0^0=1$ , the function $f(x,y)=x^y$ wouldn't be continuous in $(0,0)$ so applying limits and saying $\lim\limits_{(x,y)\to 0} x^y=0^0$ is not valid.

In Analysis it's very problematic to define $0^0$ , since we should let it be an exception to many rules. But in any way, it's necessary to define it as 1, for example, to evaluate things as: $p(x)=\displaystyle \sum_{i=0}^n a_i x^i$ in $0$ . I mean, your standard polynomial.

Or even in one of the definitions of $e^x=\displaystyle \sum_{i=0}^\infty \frac{x^i}{i!}$ we know that $e^0=1$ , and to do that we must have $0^0=1$ , since the first term of the series would be $\frac{0^0}{0!}$ .

Rewrite as

${ e }^{ \lim _{ x\rightarrow { 0 }^{ + } }{ (\ln { { x }^{ (\frac { 1 }{ lnx } ) } } ) } }$

then bring down the power on the x to the front of the natural log using the power rule, and this gives...

${ e }^{ \lim _{ x\rightarrow { 0 }^{ + } }{ (\frac { 1 }{ lnx } *(\ln { { x } } )) } }$

ln(x) statements cancel out, leaving us with

${ e }^{ \lim _{ x\rightarrow { 0 }^{ + } }{ (1) } }$

or simply, e

(ln stands for natural log)

lim x^{\frac{1}{logx}} = e^{\frac{1}{logx}] \ times (1-x) taking x=x - 1 so lim \frac{logx}{x - 1} therefore e^{1}

By definition, if $x>0$ then $x^{1/\log x}=\exp\left(\frac{1}{\log x } \cdot \log x\right)=\exp(1)=e \, .$ Since the function $x^{1/\log x}$ is identically equal to $e$ for all $x>0$ , the limit must be $e$ .

just open up: https://www.geogebra.org/calculator , and type the function. :P

Should remember the equation: x=e^lnx

So x^1/lnx= e^ln(x^1/lnx)

For log. properties ln (x^x ) = x lnx so

ln(x^1/lnx)= lnx/lnx ==> e^ln(x^1/lnx)=e

For any values of x

$x^\frac{1}{\ln x}$ is a constant function well defined in domain $\mathbb R_+$ . This value is $e$ in fact

$x^\frac{1}{\ln x} = e^{\ln\left( x^\frac{1}{\ln x} \right)} = e^{\frac{1}{\ln x} \cdot \ln x } = e^{\frac{\ln x}{\ln x}} = e^1 = e.$

So, for every $k \in \mathbb R_+ \cup \{ 0 \}$ we have

$\lim_{x \rightarrow to k} x^\frac{1}{\ln x} = \lim_{x \rightarrow k} e = e.$

I believe this can be done without the limits too (as the $lnx$ cancel out on the penultimate line):

$\lim_{x \to 0} x^\frac{1}{ln x} = \lim_{x \to 0} e^{(lnx^\frac{1}{ln x})}$

$= \lim_{x \to 0} e^{{\frac{1}{ln x}} \times lnx}$

$= \lim_{x \to 0} e^\frac{ln x}{ln x}$

$= e$

$\lim_{x \to 0^+} x^{\frac{1}{\ln{x}}} = \lim_{x \to 0^+} (e^{\ln x})^{\frac{1}{\ln{x}}} = \lim_{x \to 0^+} e^1 = e$ $\ln x$ and $\frac{1}{\ln x}$ are defined for $x > 0$ (technically excluding $1$ for $\frac{1}{\ln x}$ though that's unimportant for the limit) so we can multiply the exponents with no worries.

given the equation 0/0=x whatever value we use as x will satisfy the equation.

Let y= x^(1/log x). Thus, logy =1 Thus, limit(log y,x,0) = limit(1,x,0). By applying composite function theorem, log(limit(y,x,0) = 1. Thus, raising both terms of the equation to the power of e, we get limit(y,x,0) = e. Thus, limit(x^1/(log x) ,x,0) =e.