Math Doubts

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

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 $1$.

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

Convert Exponential term into Logarithmic form

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 $1$.

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

Transform limit of function into known form

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.

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

Use Reciprocal Limit Rule

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

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

Evaluate Limit of the Log function

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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