Question

What is the value of $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}}$ ?

Answer 1

Hint : We have to use the function change for the limit $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}}$ . The change gives the logarithm form. We have to take logarithm function on both sides of the equation $p=\underset{z\to 0}{\mathop{\lim }}\,{{\left( 1+z \right)}^{\dfrac{1}{z}}}$ . Then we use the limit formula of $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+x \right)}{x}=1$ . Using the logarithm omission, we get $p=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}}=e$ .

Complete step by step solution:
Let us take the limit as $p=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}}$ . We first interchange the variable of the given function $x=\dfrac{1}{z}$ .
The limit also changes with the change of the variable.
Therefore,

x	$\infty$
z	0

Therefore, $p=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}}=\underset{z\to 0}{\mathop{\lim }}\,{{\left( 1+z \right)}^{\dfrac{1}{z}}}$ .
Now we try to find the limit value of the function suing logarithm.
We take logarithm mon the both sides of the function $p=\underset{z\to 0}{\mathop{\lim }}\,{{\left( 1+z \right)}^{\dfrac{1}{z}}}$ .
So, \log \left( p \right)=\log \left\\{ \underset{z\to 0}{\mathop{\lim }}\,{{\left( 1+z \right)}^{\dfrac{1}{z}}} \right\\} .
We know that \log \left\\{ \underset{x\to a}{\mathop{\lim }}\,f\left( x \right) \right\\}=\underset{x\to a}{\mathop{\lim }}\,\left[ \log \left\\{ f\left( x \right) \right\\} \right] .
Therefore, \log \left( p \right)=\log \left\\{ \underset{z\to 0}{\mathop{\lim }}\,{{\left( 1+z \right)}^{\dfrac{1}{z}}} \right\\}=\underset{z\to 0}{\mathop{\lim }}\,\left[ \log \left\\{ {{\left( 1+z \right)}^{\dfrac{1}{z}}} \right\\} \right] .
Now we use the logarithm formula of $\log {{a}^{x}}=x\log a$ .
Therefore, \log \left\\{ {{\left( 1+z \right)}^{\dfrac{1}{z}}} \right\\}=\dfrac{1}{z}\log \left( 1+z \right)=\dfrac{\log \left( 1+z \right)}{z} .
The limit becomes \log \left( p \right)=\underset{z\to 0}{\mathop{\lim }}\,\left[ \log \left\\{ {{\left( 1+z \right)}^{\dfrac{1}{z}}} \right\\} \right]=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+z \right)}{z} .
We know the limit value of $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+x \right)}{x}=1$ .
Therefore, $\log \left( p \right)=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+z \right)}{z}=1$ .
Now we try to omit the logarithm of the equation $\log \left( p \right)=1$ using the formula of ${{\log }_{e}}a=y\Rightarrow a={{e}^{y}}$ .
So, $\log \left( p \right)=1$ gives $p={{e}^{1}}=e$ . This gives $p=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}}=e$ .
The value of limit $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}}$ is $e$ .
So, the correct answer is “e”.

Note : We can also directly use the limit formula of $p=\underset{z\to 0}{\mathop{\lim }}\,{{\left( 1+z \right)}^{\dfrac{1}{z}}}=e$ . The formulas $p=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}}=\underset{z\to 0}{\mathop{\lim }}\,{{\left( 1+z \right)}^{\dfrac{1}{z}}}$ are the derivation of two formulas for one another.