Courses
Courses for Kids
Free study material
Free LIVE classes
More LIVE
Join Vedantu’s FREE Mastercalss

# Evaluate $\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{{{(1+\left[ x \right])}^{\dfrac{1}{\left\{ x \right\}}}}}{e} \right)}^{\dfrac{1}{\left\{ x \right\}}}}$if it exist (where $\left\{ x \right\}$ denotes the fractional part of x). Verified
362.7k+ views
Hint: Convert fractional part function to greatest integer function and solve by substituting $x$ as $\left( 0+h \right)$ or $\left( 0-h \right)$.

Consider the given expression,
$\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{{{(1+\left[ x \right])}^{\dfrac{1}{\left\{ x \right\}}}}}{e} \right)}^{\dfrac{1}{\left\{ x \right\}}}}$
Here $\left\{ x \right\}$ denotes the fractional part of x.
We know fractional part will always be non-negative and fractional part is greater than or equal to $'0'$ and less than $'1'$ .
Here in the given equation, we can apply the formula,
$\underset{x\to 0}{\mathop{\lim }}\,{{(1+x)}^{\dfrac{1}{x}}}=\underset{x\to 0}{\mathop{\lim }}\,\text{ e}$
Now, simplifying the given expression, we get
$\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{{{(1+\left[ x \right])}^{\dfrac{1}{\left\{ x \right\}}}}}{e} \right)}^{\dfrac{1}{\left\{ x \right\}}}}=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{e}{e} \right)}^{\dfrac{1}{\left\{ x \right\}}}}$
Cancelling the like terms, we get
$\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{{{(1+\left[ x \right])}^{\dfrac{1}{\left\{ x \right\}}}}}{e} \right)}^{\dfrac{1}{\left\{ x \right\}}}}=\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1 \right)}^{\dfrac{1}{\left\{ x \right\}}}}...........(i)$
We know the expansion,
${{a}^{x}}=1+\dfrac{x\ln a}{1!}+\dfrac{{{x}^{2}}{{\ln }^{2}}a}{2!}+.....$
Applying this in equation (i), we get
$\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{{{(1+\left[ x \right])}^{\dfrac{1}{\left\{ x \right\}}}}}{e} \right)}^{\dfrac{1}{\left\{ x \right\}}}}=\underset{x\to 0}{\mathop{\lim }}\,\left( 1+\dfrac{\left\{ x \right\}\ln (1)}{1!}+\dfrac{{{\left\{ x \right\}}^{2}}{{\ln }^{2}}(1)}{2!}+..... \right)$
But we know, $\ln 1=0$ , so above equation becomes,
$\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{{{(1+\left[ x \right])}^{\dfrac{1}{\left\{ x \right\}}}}}{e} \right)}^{\dfrac{1}{\left\{ x \right\}}}}=\underset{x\to 0}{\mathop{\lim }}\,\left( 1+\dfrac{0}{1!}+\dfrac{0}{2!}+..... \right)$
As we can see that the limit is free from $'x'$ term. So the limit of the function will be constant term at any point. So we get
$\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{{{(1+\left[ x \right])}^{\dfrac{1}{\left\{ x \right\}}}}}{e} \right)}^{\dfrac{1}{\left\{ x \right\}}}}=1$

Note: Students usually don’t learn expansions and are struck while solving the questions.
See the fractional part the student think it is very difficult.
They start applying,
$x=[x]+\{x\}$
$\therefore \{x\}=x-[x]$
And substitute this in the given expression, leading to more confusion and ending up in wrong answer.
Last updated date: 19th Sep 2023
Total views: 362.7k
Views today: 8.62k