Question

What is the limit of \[{{\left( 1+x \right)}^{\dfrac{1}{x}}}\] as \[x\] approaches \[0\]?

Answer 1

Hint: In order to find the limit of \[{{\left( 1+x \right)}^{\dfrac{1}{x}}}\] when \[x\] approaches \[0\], firstly, since it is a binomial expansion, we have to check out the general expansion of given binomial expression and then after expanding it accordingly and then applying the limits gives us the required answer.

Complete step by step answer:
Let us see the general result of a limit when it approaches zero. The limit of \[f\left( x \right)\] as \[x\] approaches \[0\] is undefined, since both sides approach different values. When simply evaluating an equation \[0/0\] is undefined. However, if we take the limit, if we get \[0/0\] we can get a variety of answers and the only way to know which one is correct is to actually compute the limit.
So let us start computing the limit of our given function.
Firstly, consider the binomial expansion,
\[{{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+...\]
From the above expansion, we have
\[{{\left( 1+x \right)}^{\dfrac{1}{x}}}=1+\dfrac{1}{x}x+\dfrac{\dfrac{1}{x}\left( \dfrac{1}{x}-1 \right)}{2!}{{\dfrac{1}{x}}^{2}}+\dfrac{\dfrac{1}{x}\left( \dfrac{1}{x}-1 \right)\left( \dfrac{1}{x}-2 \right)}{3!}{{\dfrac{1}{x}}^{3}}+...\]
Upon solving it, we get
\[=1+1+\dfrac{1\left( 1-x \right)}{2!}+\dfrac{1\left( 1-x \right)\left( 1-2x \right)}{3!}+...\]
Now let us apply the limits to the expansion
\[\displaystyle \lim_{x \to 0}{{\left( 1+x \right)}^{\dfrac{1}{x}}}=\displaystyle \lim_{x \to 0}1+1+\dfrac{1\left( 1-x \right)}{2!}+\dfrac{1\left( 1-x \right)\left( 1-2x \right)}{3!}+...\]
Upon evaluating the limits, we get
\[\Rightarrow \sum\limits_{k=0}^{\infty }{\dfrac{1}{k!}=e}\]
\[\therefore \] The limit of the function \[{{\left( 1+x \right)}^{\dfrac{1}{x}}}\] as \[x\] approaches \[0\] is \[e\].

Note: We can apply L-hospital rule if the function is indeterminate form \[0/0\] or \[\infty /\infty \] all we need to do is differentiate the numerator and differentiate the denominator and then take the limit. L'Hopital's rule only applies when the expression is indeterminate, i.e. \[0/0\] or \[\infty /\infty \]. So we should not apply the rule when we have a determinable form.