Question

How do you find the limit of ${\left( {x - \dfrac{1}{x}} \right)^x}$ as $x$ approaches to infinity ?

Answer 1

As we know that the above question is an example of limit function. A limit of a function can be defined as a number that a function reaches as the independent variable of the function reaches at a given number. The limit of a real valued function ”$f$” with respect to the variable $x$ can be defined as the $\mathop {\lim }\limits_{x \to p} f(x) = L$. Here lim refers to the limit of the function. We can say that the limit of any given function $'f'$ of $x$ as $x$ approaches to $p$ is equal to $L$.

Complete step by step solution:
We can write the given function as $\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \dfrac{1}{x}} \right)^x}$, which has the form ${1^\infty }$ as it is an intermediate form. We can use the
logarithms and exponential function to solve this. We that if any function is limit approaches to infinity it can be written as in $'e'$ form. So here we have ${\left( {1 - \dfrac{1}{x}} \right)^x} = {e^{1n{{\left( {1 - \dfrac{1}{x}} \right)}^x}}}$.
We can now solve for the limit of the exponent: $\mathop {\lim }\limits_{x \to \infty } \left( {1n{{\left( {1 - \dfrac{1}{x}} \right)}^x}} \right)$ , we can write $1 - \dfrac{1}{x}$ as $\dfrac{{x - 1}}{x}$.
So $1n{\left( {1 - \dfrac{1}{x}} \right)^x} = 1n{\left( {\dfrac{{x - 1}}{x}} \right)^x}$. Here by using the property of logarithms , the power will come in the front i.e. $xIn\left( {\dfrac{{x - 1}}{x}} \right)$. Now as $x \to \infty$ so $In1 = \infty$. By using the Hopital’s rule we will put the
reciprocal of these in the denominator: $xIn\left( {\dfrac{{x - 1}}{x}} \right) = \dfrac{{In\left( {\dfrac{{x - 1}}{x}} \right)}}{{\dfrac{1}{x}}},x \to \infty$ , we get the form of $\dfrac{0}{0}$ to apply L'hopital's rule: $\dfrac{d}{{dx}}\left( {In\left( {\dfrac{{x - 1}}{x}} \right)} \right) = \dfrac{1}{{\dfrac{{x - 1}}{x}}} \times \dfrac{d}{{dx}}\left( {\dfrac{{x - 1}}{x}} \right)$, It can further be solved $\dfrac{x}{{x - 1}} \times \dfrac{1}{{{x^2}}} = \dfrac{1}{{x(x - 1)}}$.
So we get the form
$\dfrac{{1n\left( {\dfrac{{x - 1}}{x}} \right)}}{{\dfrac{1}{x}}} = \dfrac{{\dfrac{1}{{x(x - 1)}}}}{{\dfrac{1}{x}}}$, It gives $- \dfrac{x}{{x - 1}}$.
Now solve with the limit: $\mathop {\lim }\limits_{x \to \infty } \left( { - \dfrac{x}{{x - 1}}} \right) = - 1$ $\Rightarrow {\left( {1 - \dfrac{1}{x}} \right)^x} = {e^{In{{\left( {1 - \dfrac{1}{x}} \right)}^x}}}$,
As $x \to \infty$ the exponent goes to $- 1$. Therefore $\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \dfrac{1}{x}} \right)^x} = {e^{ - 1}} = \dfrac{1}{e}$.
Hence the required answer is $\dfrac{1}{e}$.

Note: As we know that according to L’Hopital’s rule, the limit of a function when we divide one function by another function, it remains the same after we take the derivative of each function. The number $e$ is also known as Euler’s number which is a mathematical constant. It is also the base of the natural logarithm. It is defined by $e = \mathop {\lim }\limits_{n \to \infty } (1 + 1n)n$