Question

Show that \(\mathop {\lim }\limits_{n \to \infty } {n^4}{e^{ - n}} = 0{\rm{ }}\)\({\rm{and use a graph to find the smallest value of}}\)N \({\rm{\;that corresponds to}}\) \({\rm{\varepsilon = 0}}{\rm{.1}}\) \({\rm{to\;\;in the precise definition of a limit}}{\rm{.}}\)

Short answer

The smallest value of N should be \( \ge 12.36\)

Step-by-step solution

 Definition of limit

A limit is a value that a function approaches for the specified input values as it approaches the output. Limits are used to define integrals, derivatives, and continuity in calculus and mathematical analysis.

Finding the indeterminate form using L Hospital’s rule

Let us take\(S = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}}\)

The above expression has indeterminate\((\frac{\infty }{\infty }{\rm{)}}\)

We know that for an indeterminate form one can apply L'Hospital's Rule given as

\(\mathop {\lim }\limits_{n \to \infty } \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{f^\prime }(x)}}{{{g^\prime }(x)}}\)

Therefore, applying L'Hospital's Rule four times, we have

\(\begin{array}{c}S = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}}\\\mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}} = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{4{n^3}}}{{{e^n}}}} \right];\quad ({\rm{applying L'Hospital's Rulefirsttime}})\\\mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}} = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{12{n^2}}}{{{e^n}}}} \right];\quad ({\rm{applying L'Hospital's Rulesecondtime}})\\\mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}} = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{24n}}{{{e^n}}}} \right];\quad ({\rm{applying L'Hospital's Rulethirdtime}})\\\mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}} = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{24}}{{{e^n}}}} \right];\\\mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}} = 24\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{e^n}}}} \right]({\rm{applying L'Hospital's Rule fourth time}})\end{array}\)

\(\begin{array}{c}\mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}} = 24 \cdot \left[ {\frac{1}{{{e^\infty }}}} \right]\\\mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}} = 24 \cdot 0\\\mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}} = 0\\\mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}} = 0\end{array}\)

Finding the smallest value of N

In the precise definition of a limit, we have

\(\mathop {\lim }\limits_{x \to \infty } f(x) = L\)

if for every number\(\varepsilon \)>0, there is some number N>0

such that

\(|f(x) - L| < {\rm{whenever }}x > N\)

Therefore, we have

\(\begin{array}{*{20}{r}}{\left| {\frac{{{n^4}}}{{{e^n}}} - 0} \right| < 0.1}\\{\left| {\frac{{{n^4}}}{{{e^n}}}} \right| < 0.1}\\{\left( {\frac{{{n^4}}}{{{e^n}}}} \right) < 0.1}\end{array}\)

whenever n>N

\(We\;plot\left( {\frac{{{n^4}}}{{{e^n}}}} \right)vs\;n\;in\;the\;given\;figure.\;And\;we\;find\;out\;that\;for\)

\(\left( {\frac{{{n^4}}}{{{e^n}}}} \right) < 0.1\quad {\rm{ for }}n > 12.36\)

And we find out that the smallest value of N that corresponds to\(\varepsilon \)=0.1 in the precise definition of a limit is

\(\begin{array}{l}N = 12.36\\N \ge 12.36\end{array}\)

The smallest value of N is

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{n^4}}}{{{e^n}}} = 0.1\)

,where

\(N \ge 12.36\)