An exponential term and natural logarithmic function are involved in algebraic function in this limit problem and the limit of this exponential logarithmic algebraic function has to evaluated as the variable $x$ approaches $1$.

$\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^x-1}{x\ln{x}}}$

Transform the exponential term into logarithmic term by using relation between exponents and logarithms.

Take $p = x^x$, then $\ln{p} = \ln{x^x}$. Therefore, $\ln{p} = x\ln{x}$ as per power rule of logarithms. Now, replace $x^x$ by $p$ and also replace $x\ln{x}$ by $\ln{p}$ in the algebraic fractional function.

According to $p = x^x$, If $x \to 1$, then $x^x \to 1^1$, Therefore, $x^x \to 1$ but $x^x$ is taken as $p$. Therefore, if $x$ approaches $1$ then $p$ also approaches $1$.

$= \,\,\,$ $\displaystyle \large \lim_{p \,\to\, 1}{\normalsize \dfrac{p-1}{\ln{p}}}$

The limit of the algebraic function has to be simplified further to transform it into known form for evaluating the limit of this function.

If $p \to 1$ then $p-1 \to 0$. In other words, If $p$ tends to $1$, then $p-1$ approaches to $0$.

$= \,\,\,$ $\displaystyle \large \lim_{p-1 \,\to\, 0}{\normalsize \dfrac{p-1}{\ln{p}}}$

Now, take $k = p-1$, then $p = k+1$

Now, convert the total function in terms of $k$ by the relation between variables $k$ and $p$.

$= \,\,\,$ $\displaystyle \large \lim_{k \,\to\, 0}{\normalsize \dfrac{k}{\ln{(k+1)}}}$

$= \,\,\,$ $\displaystyle \large \lim_{k \,\to\, 0}{\normalsize \dfrac{k}{\ln{(1+k)}}}$

This mathematical expression is similar to the limit rule of logarithmic function but it should be in reciprocal form. So, express this algebraic function into its reciprocal form.

$= \,\,\,$ $\displaystyle \large \lim_{k \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\ln{(1+k)}}{k}}}$

According to reciprocal rule of limits, the limit of the reciprocal function can be written as follows.

$= \,\,\,$ $\dfrac{1}{\displaystyle \large \lim_{k \,\to\, 0}{\normalsize \dfrac{\ln{(1+k)}}{k}}}$

The denominator of the fraction is completely in the form of limit rule of logarithmic function. So, it can be used to the limit of the function as $k$ approaches zero.

$= \,\,\,$ $\dfrac{1}{1}$

$= \,\,\,$ $1$

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