Question

Find the limit of $\underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{r=2}^{n}{\left( \dfrac{{{r}^{3}}+1}{{{r}^{3}}-1} \right)}$ =
1. $1$
2. $\dfrac{1}{2}$
3. $\dfrac{3}{2}$
4. $2$

Answer 1

Hint : To solve this question first start by simplifying the bracket using the cube formula to make it a little easier to begin solving. Then the student needs to find the product of the expression from the range of 2 to infinity. Once we simplify and solve the product part then the student will find the limit of the function left at the point where n tends to infinity.

Complete step-by-step answer :
Now the expression needed to solve is
$\underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{r=2}^{n}{\left( \dfrac{{{r}^{3}}+1}{{{r}^{3}}-1} \right)}$
Simplifying the brackets and its numerator and denominator using the cube formula we get
$\underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{r=2}^{n}{\left( \dfrac{{{r}^{3}}+{{1}^{3}}}{{{r}^{3}}-{{1}^{3}}} \right)}$
Now using the formula of ${{r}^{3}}+{{1}^{3}}=\left( r+1 \right)\left( {{r}^{2}}-r+1 \right)$ and ${{r}^{3}}-{{1}^{3}}=\left( r-1 \right)\left( {{r}^{2}}+r+1 \right)$
$\underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{r=2}^{n}{\dfrac{\left( r+1 \right)\left( {{r}^{2}}-r+1 \right)}{\left( r-1 \right)\left( {{r}^{2}}+r+1 \right)}}$
Now taking the products individually of first bracket and second bracket
$\underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{r=2}^{n}{\dfrac{\left( r+1 \right)}{\left( r-1 \right)}}\times \underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{r=2}^{n}{\dfrac{\left( {{r}^{2}}-r+1 \right)}{\left( {{r}^{2}}+r+1 \right)}}$
Now therefore simplifying and taking the products we get
\underset{n\to \infty }{\mathop{\lim }}\,\left\\{ \dfrac{3}{1}.\dfrac{4}{2}.\dfrac{5}{3}.\dfrac{6}{4}.....\dfrac{n-2}{n-5}.\dfrac{n-1}{n-3}\dfrac{n}{n-2}.\dfrac{n+1}{n-1} \right\\}\times \underset{n\to \infty }{\mathop{\lim }}\,\left\\{ \dfrac{3}{7}.\dfrac{7}{13}.\dfrac{13}{21}......\dfrac{{{n}^{2}}-5n+7}{{{n}^{2}}-3n+3}.\dfrac{{{n}^{2}}-3n+3}{{{n}^{2}}-n+1}.\dfrac{{{n}^{2}}-n+1}{{{n}^{2}}+n+1} \right\\}
Now to get the simplified answer here we cancel the same terms so as we can see in first part the numerator of the second term is same as denominator of fourth term. So we use the same thing to cancel all terms in between which will lead us with the fourth last term of numerator being the same as that of second last of denominator. Hence we can cancel all terms in between.
Similarly in second part of products you can see that the denominator of first term is same as numerator of last one and in the same pattern it goes to numerator of last one being the same as denominator or second last and we can cancel anything in between.
\underset{n\to \infty }{\mathop{\lim }}\,\left\\{ \dfrac{3n(n+1)}{1.2({{n}^{2}}+n+1)} \right\\}
Now taking the constant part of this equation out of limits
\underset{n\to \infty }{\mathop{\dfrac{3}{2}\lim }}\,\left\\{ \dfrac{n(n+1)}{({{n}^{2}}+n+1)} \right\\}
Now to get the limit we divide both the numerator and denominator by the square of n by doing that we simplify it and we get
\underset{n\to \infty }{\mathop{\dfrac{3}{2}\lim }}\,\left\\{ \dfrac{\dfrac{n(n+1)}{{{n}^{2}}}}{(\dfrac{{{n}^{2}}}{{{n}^{2}}}+\dfrac{n}{{{n}^{2}}}+\dfrac{1}{{{n}^{2}}})} \right\\}
Now in numerator dividing both terms by n and in denominator dividing all three terms by square of n we get
\underset{n\to \infty }{\mathop{\dfrac{3}{2}\lim }}\,\left\\{ \dfrac{1+\dfrac{1}{n}}{1+\dfrac{1}{n}+\dfrac{1}{{{n}^{2}}}} \right\\}
Now to find the limit at the point where n tends to infinity we puts n value in the function and we get
$\dfrac{3}{2}\times \dfrac{1+0}{1+0+0}$
Now simplifying this we get limit of the expression therefore
$=\dfrac{3}{2}$
Hence we can say that the limit of the expression is
$\underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{r=2}^{n}{\left( \dfrac{{{r}^{3}}+1}{{{r}^{3}}-1} \right)}=\dfrac{3}{2}$
The answer for the question therefore is option 3
So, the correct answer is “Option 3”.

Note : To explain what limit is in mathematics in simple words we can say that a limit is the value that the function or sequence or expression approaches as input or we can also say index tends to some value. A common mistake made is students directly apply the value of limit which leaves us with $\dfrac{\infty }{\infty }$ which is mathematically incorrect.