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.

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

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

Latest Math Topics

Mar 21, 2023

Feb 25, 2023

Feb 17, 2023

Feb 10, 2023

Latest Math Problems

Mar 03, 2023

Mar 01, 2023

Feb 27, 2023

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the math problems in different methods with understandable steps and worksheets on every concept for your practice.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved