# Evaluate $\displaystyle \large \lim_{x \,\to\, e}{\normalsize \dfrac{\log_{e}{x}-1}{x-e}}$

$\dfrac{\log_{e}{x}-1}{x-e}$ is an algebraic function in which a natural logarithmic term is involved. It’s our job to find the limit of this function as $x$ approaches zero.

$\displaystyle \large \lim_{x \,\to\, e}{\normalsize \dfrac{\log_{e}{x}-1}{x-e}}$

### Basic steps to convert function

When a function is in logarithmic form in limits, you must consider properties of limits for the logarithmic functions. Let us take some basic steps to transform the whole function same as our standard result in the limits.

If $x \,\to\, e$, then $x-e \,\to\, 0$. Therefore, if $x$ approaches $e$, then $x-e$ approaches $0$.

$= \,\,\,$ $\displaystyle \large \lim_{x -e \,\to\, 0}{\normalsize \dfrac{\log_{e}{x}-1}{x-e}}$

Now, take $x-e = u$, then $x = u+e$. Now, convert the entire function in terms of $u$ from $x$.

$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\log_{e}{(u+e)}-1}{u}}$

### Use Logarithmic identities for simplification

Let’s simplify the expression in the numerator by using properties of logarithms.

$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\log_{e}{(u+e)}-1}{u}}$

The first term is in natural logarithm form but the second term is a number. If the number $1$ is written in term of natural logarithm, then the terms in the numerator can be merged. It’s possible to write $1$ as $\ln{(e)}$ or $\log_{e}{(e)}$ as per log of base rule.

$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\log_{e}{(u+e)}-\log_{e}{e}}{u}}$

There are two natural logarithmic terms in the numerator and they can be merged by using quotient rule of logarithms.

$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\log_{e}{ \Big(\dfrac{u+e}{e}\Big)}}{u}}$

Now, continue simplifying the logarithmic function in the numerator further.

$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(\dfrac{u}{e}+\dfrac{e}{e}\Big)}}{u}}$

$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(\dfrac{u}{e}+\dfrac{\cancel{e}}{\cancel{e}}\Big)}}{u}}$

$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(\dfrac{u}{e}+1\Big)}}{u}}$

$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{u}{e}\Big)}}{u}}$

### Transform the function to Standard Result

Take $m = \dfrac{u}{e}$, then $u = me$.

If, $u \,\to\, 0$, then $\dfrac{u}{e} \,\to\, \dfrac{0}{e}$. Therefore, $\dfrac{u}{e} \,\to\, 0$ but $\dfrac{u}{e} = m$. So, $m \,\to\, 0$. It’s derived mathematically that if $u$ approaches $0$, then $m$ also approaches to $0$.

Now, transform the whole limit function in terms of $m$ from $u$.

$= \,\,\,$ $\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+m)}}{me}}$

We can continue the simplification of the function.

$= \,\,\,$ $\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+m)}}{m \times e}}$

$= \,\,\,$ $\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \Bigg[ \dfrac{1}{e} \times \dfrac{\log_{e}{(1+m)}}{m} \Bigg]}$

$= \,\,\,$ $\dfrac{1}{e} \times \displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+m)}}{m}}$

### Find the Limit of the Logarithmic function

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

We know that the limit of ln(1+x)/x as x approaches 0 is equal to one. Similarly, the limit of $\ln{(1+m)}/m$ as $m$ tends to $0$ must also be equal to $1$.

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

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

In this limit problem, it has proved that the limit of function $\dfrac{\log_{e}{x}-1}{x-e}$ as $x$ approaches zero is equal to $\dfrac{1}{e}$

Email subscription
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
###### Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more