# 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).

Last updated date: 18th Mar 2023

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Answer

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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.

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.

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