Question

$\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}{{\left( 1-\dfrac{m}{n} \right)}^{n-x}}$is equal to
( a ) 0
( b ) 1
( c ) $\dfrac{{{m}^{x}}.{{e}^{-m}}}{x!}$
( d ) $\dfrac{{{m}^{x}}.{{e}^{m}}}{x!}$

Answer 1

In this question what we have asked to do is we have to evaluate the given limit $\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}{{\left( 1-\dfrac{m}{n} \right)}^{n-x}}$. What we will do is we will solve this question using the formula of binomial expansion and using its properties and then we apply the limit to evaluate the value of the function.

Complete step by step answer:
In given question, we have the function $\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}{{\left( 1-\dfrac{m}{n} \right)}^{n-x}}$. We want to find the limit of function as$n\to \infty$. so, what we will do is, we will first simplify the given function into simpler component functions and then apply the limit to evaluate the value of the function.
We know the combinatorial formula${}^{n}{{C}_{x}}=\dfrac{n!}{x!\left( n-x \right)!}$ and $n!=n(n-1)(n-2)(n-3)......3\cdot 2\cdot 1$ known as factorial function
This can be simplified by cutting out the like terms in the numerator and denominator of the above formula to get${}^{n}{{C}_{x}}=\dfrac{n!}{x!\left( n-x \right)!}=\dfrac{n\left( n-1 \right)\left( n-2 \right)...3\cdot 2\cdot 1}{x!}$
Or, ${}^{n}{{C}_{x}}=\dfrac{n!}{x!\left( n-x \right)!}=\dfrac{n\left( n-1 \right)\left( n-2 \right)...\left( n-x+1 \right)(n-x)!}{x!(n-x)!}$
On simplifying we get,
${}^{n}{{C}_{x}}=\dfrac{n!}{x!\left( n-x \right)!}=\dfrac{n\left( n-1 \right)\left( n-2 \right)...\left( n-x+1 \right)}{x!}$…..( i)
Now, we will consider the function${}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}$. We can simplify and write ${}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}$ as using equation ( i)
${}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)...\left( n-x+1 \right)}{x!}\left( \dfrac{{{m}^{x}}}{{{n}^{x}}} \right)$……( ii )
Taking n common from all terms in numerator, we get
${}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}=\dfrac{{{n}^{x}}\left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)...\left( 1-\dfrac{x+1}{n} \right)}{x!}\left( \dfrac{{{m}^{x}}}{{{n}^{x}}} \right)$
Now, we will divide both numerator and denominator by${{n}^{x}}$to simplify the function and get
$=\dfrac{1\left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)...\left( 1+\dfrac{1-x}{n} \right)}{x!}{{m}^{x}}$
If we apply limit to this function as$n\to \infty$, we will get
$\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{n\left( n-1 \right)\left( n-2 \right)...\left( n-x+1 \right)}{x!}\left( \dfrac{{{m}^{x}}}{{{n}^{x}}} \right)$
$\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1\left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)...\left( 1+\dfrac{1-x}{n} \right)}{x!}{{m}^{x}}$.
We observe that as$n\to \infty$, we get$\dfrac{1}{n}\to 0$.
Hence, all the terms in the numerator of$\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1\left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)...\left( 1+\dfrac{1-x}{n} \right)}{x!}{{m}^{x}}$achieves the value$1$.
Thus, we get
$\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{n\left( n-1 \right)\left( n-2 \right)...\left( n-x+1 \right)}{x!}\left( \dfrac{{{m}^{x}}}{{{n}^{x}}} \right)$
$\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1\left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)...\left( 1+\dfrac{1-x}{n} \right)}{x!}{{m}^{x}}=1\dfrac{{{m}^{x}}}{x!}$
$\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}=\dfrac{{{m}^{x}}}{x!}$.
Now, we observe the term${{\left( 1-\dfrac{m}{n} \right)}^{n-x}}$. As we apply the limits, we observe that we can write${{\left( 1-\dfrac{m}{n} \right)}^{n-x}}$as
${{\left( 1-\dfrac{m}{n} \right)}^{n-x}}=1-\left( \dfrac{m}{n} \right)+{{\left( \dfrac{m}{n} \right)}^{2}}\dfrac{1}{2!}-{{\left( \dfrac{m}{n} \right)}^{3}}\dfrac{1}{3!}+...$.
Now, see if $n\to \infty$, then in ${{\left( 1-\dfrac{m}{n} \right)}^{n-x}}$, the value inside bracket tends to 1 and power tends to$\to \infty$which form case of ${{1}^{\infty }}$ which is inderminate form.
And we know that $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( f(x) \right)}^{g(x)}}={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,g(x)\log f(x)}}$ , for ${{1}^{\infty }}$case.
Hence, we can write$\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1-\dfrac{m}{n} \right)}^{n-x}}=\underset{n\to \infty }{\mathop{\lim }}\,{{e}^{(n-x)\cdot log\left( 1-\dfrac{m}{n} \right)}}$.
We know that, $\log (1-x)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-........$
So, $\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1-\dfrac{m}{n} \right)}^{n-x}}=\underset{n\to \infty }{\mathop{\lim }}\,{{e}^{(n-x)\cdot \left( -\dfrac{m}{n}-\dfrac{{{\left( \dfrac{m}{n} \right)}^{2}}}{2}-\dfrac{{{\left( \dfrac{m}{n} \right)}^{3}}}{3}-\dfrac{{{\left( \dfrac{m}{n} \right)}^{4}}}{4}-........ \right)}}$
On solving we get,
$\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1-\dfrac{m}{n} \right)}^{n-x}}={{e}^{-m}}$
We have$\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1-\dfrac{m}{n} \right)}^{n-x}}={{e}^{-m}}$and$\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}=\dfrac{{{m}^{x}}}{x!}$.
We know that, $\underset{x\to a}{\mathop{\lim }}\,f(x)\cdot g(x)=\underset{x\to a}{\mathop{\lim }}\,f(x)\cdot \underset{x\to a}{\mathop{\lim }}\,g(x)$ , so using this property
Thus, by combining the two formulas, we get
$\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}{{\left( 1-\dfrac{m}{n} \right)}^{n-x}}=\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1-\dfrac{m}{n} \right)}^{n-x}}$
$\underset{n\to \infty }{\mathop{\lim }}\,{}^{n}{{C}_{x}}{{\left( \dfrac{m}{n} \right)}^{x}}{{\left( 1-\dfrac{m}{n} \right)}^{n-x}}=\dfrac{{{m}^{x}}}{x!}\left( {{e}^{-m}} \right)=\dfrac{{{m}^{x}}{{e}^{-m}}}{x!}$as the required limit.

So, the correct answer is “Option C”.

Note: It’s necessary to first simplify the expressions and then apply the limit into component functions instead of directly applying the limit to the actual function. We can see that even after applying the limit$n\to \infty$, it’s not necessary that the function will blow up to$\infty$. We can get a finite value of the function even after applying an infinite limit.