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.