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Question: How do you determine the limit of \({{\left( \dfrac{n+5}{n-1} \right)}^{n}}\) as n approaches to \(\...

How do you determine the limit of (n+5n−1)n{{\left( \dfrac{n+5}{n-1} \right)}^{n}} as n approaches to ∞\infty ?

Explanation

Solution

In this question, we have to find the limit of a fractional algebraic expression. Thus, we will apply the limit formula and the basic mathematical rule to get the solution. First, we will apply the exponential-logarithm formula x=elog⁡xx={{e}^{\log x}} in the function, then we will apply the log formula log⁡ab=blog⁡a\log {{a}^{b}}=b\log a. After that, we will put the value of limit in the problem, thus get the indeterminate form, therefore, we will apply the l’Hopital rule in the expression and make the necessary calculations to get the required solution for the problem.

Complete step-by-step solution:
According to the question, we have to find the limit of a fractional algebraic expression.
Thus, we will use the limit formula and the basic mathematical rule to get the solution.
The expression given to us is (n+5n−1)n{{\left( \dfrac{n+5}{n-1} \right)}^{n}} where n tends to ∞\infty --------- (1)
First, we will write the equation (1) in terms of limit, we get
lim⁡n→∞(n+5n−1)n\displaystyle \lim_{n \to \infty }{{\left( \dfrac{n+5}{n-1} \right)}^{n}}
Now, we will apply the exponential-logarithm formula x=elog⁡xx={{e}^{\log x}} in the above limit, we get
⇒lim⁡n→∞elog⁡(n+5n−1)n\Rightarrow \displaystyle \lim_{n \to \infty }{{e}^{\log {{\left( \dfrac{n+5}{n-1} \right)}^{n}}}}
Now, we will apply the logarithm formula log⁡ab=blog⁡a\log {{a}^{b}}=b\log a in the above limit, we get
⇒lim⁡n→∞enlog⁡(n+5n−1)\Rightarrow \displaystyle \lim_{n \to \infty }{{e}^{n\log \left( \dfrac{n+5}{n-1} \right)}}
On further solving, we get
⇒lim⁡n→∞elog(n+5n−1)1n\Rightarrow \displaystyle \lim_{n \to \infty }{{e}^{\dfrac{log\left( \dfrac{n+5}{n-1} \right)}{\dfrac{1}{n}}}}
Now, we will apply the limit formula lim⁡ex=elim⁡x\lim {{e}^{x}}={{e}^{\lim x}} in the above expression, we get
⇒elim⁡n→∞(log(n+5n−1)1n)\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{log\left( \dfrac{n+5}{n-1} \right)}{\dfrac{1}{n}} \right)}}
As we see from the above equation that when we put the limit of n in the expression, we get 00\dfrac{0}{0} form, therefore we will apply the l’Hopital rule in the above expression. That is, we will differentiate the numerator with respect to n and differentiate the denominator with respect to n, we get
⇒elim⁡n→∞(ddnlog(n+5n−1)ddn1n)\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{\dfrac{d}{dn}log\left( \dfrac{n+5}{n-1} \right)}{\dfrac{d}{dn}\dfrac{1}{n}} \right)}}
On solving the above differentiation, we get
⇒elim⁡n→∞(1(n+5n−1).(n−1)−(n+5)(n−1)2(−1n2))\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{\dfrac{1}{\left( \dfrac{n+5}{n-1} \right)}.\dfrac{\left( n-1 \right)-\left( n+5 \right)}{{{\left( n-1 \right)}^{2}}}}{\left( \dfrac{-1}{{{n}^{2}}} \right)} \right)}}
On further simplification, we get
⇒elim⁡n→∞(−n21n+5.n−1−n−5(n−1))\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( -{{n}^{2}}\dfrac{1}{n+5}.\dfrac{n-1-n-5}{\left( n-1 \right)} \right)}}
⇒elim⁡n→∞(−n21n+5.−6(n−1))\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( -{{n}^{2}}\dfrac{1}{n+5}.\dfrac{-6}{\left( n-1 \right)} \right)}}
⇒elim⁡n→∞(6n2(n+5)(n−1))\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{6{{n}^{2}}}{\left( n+5 \right)\left( n-1 \right)} \right)}}
On putting the limit in the above expression, we again get the indeterminate form ∞∞\dfrac{\infty }{\infty } , thus we will use the l’Hopital rule, that is That is, we will differentiate the numerator with respect to n and differentiate the denominator with respect to n, we get
⇒elim⁡n→∞(6(2n)(n−1)+(n+5))\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{6\left( 2n \right)}{\left( n-1 \right)+\left( n+5 \right)} \right)}}
Therefore, we get
⇒elim⁡n→∞(6(2n)n−1+n+5)\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{6\left( 2n \right)}{n-1+n+5} \right)}}
⇒elim⁡n→∞(6(2n)2n+4)\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{6\left( 2n \right)}{2n+4} \right)}}
Now, we will take the common 2 from the denominator in the power, we get
⇒elim⁡n→∞(6(2n)2(n+2))\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{6\left( 2n \right)}{2\left( n+2 \right)} \right)}}
On further solving the above expression, we get
⇒elim⁡n→∞(6(n)(n+2))\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{6\left( n \right)}{\left( n+2 \right)} \right)}}
So, we will divide n on the numerator and the denominator in the above expression, we get
⇒elim⁡n→∞(6(nn)(n+2)n)\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{6\left( \dfrac{n}{n} \right)}{\dfrac{\left( n+2 \right)}{n}} \right)}}
Therefore, we get
⇒elim⁡n→∞(61+2n)\Rightarrow {{e}^{\displaystyle \lim_{n \to \infty }\left( \dfrac{6}{1+\dfrac{2}{n}} \right)}}
Now, we will again apply the limit formula lim⁡ab=lim⁡alim⁡b\lim \dfrac{a}{b}=\dfrac{\lim a}{\lim b} in the above expression, we get
⇒e(lim⁡n→∞6lim⁡n→∞(1+2n))\Rightarrow {{e}^{\left( \dfrac{\displaystyle \lim_{n \to \infty }6}{\displaystyle \lim_{n \to \infty }\left( 1+\dfrac{2}{n} \right)} \right)}}
So, we will again apply the limit formula lim⁡n→∞a=alim⁡n→∞1=a\displaystyle \lim_{n \to \infty }a=a\displaystyle \lim_{n \to \infty }1=a in the numerator and lim⁡n→∞(a+n)=alim⁡n→∞1+lim⁡n→∞n\displaystyle \lim_{n \to \infty }\left( a+n \right)=a\displaystyle \lim_{n \to \infty }1+\displaystyle \lim_{n \to \infty }n in the denominator, we get
⇒e(6lim⁡n→∞1+lim⁡n→∞2n)\Rightarrow {{e}^{\left( \dfrac{6}{\displaystyle \lim_{n \to \infty }1+\displaystyle \lim_{n \to \infty }\dfrac{2}{n}} \right)}}
Therefore, we get
⇒e(61+2∞)\Rightarrow {{e}^{\left( \dfrac{6}{1+\dfrac{2}{\infty }} \right)}}
As we know, 2∞=0\dfrac{2}{\infty }=0 , thus we get
⇒e(61+0)\Rightarrow {{e}^{\left( \dfrac{6}{1+0} \right)}}
⇒e6\Rightarrow {{e}^{6}} which is the required answer.
Therefore, the limit of (n+5n−1)n{{\left( \dfrac{n+5}{n-1} \right)}^{n}} as n approaches to ∞\infty is equal to e6{{e}^{6}}.

Note: While solving this problem, do mention each step properly to avoid confusion and mathematical error. Always remember that when applying the l’Hopital rule, you have to do the differentiation separately, that is do not apply the differentiation in the division formula, instead do the differentiation in the numerator and the denominator separately.