Citeprep Guides

When

a_n

is a quotient of two functions

111800:front

Given

Find

CitePrep Guides

Problem No. 2

HOW

Problem No. 2

HOW

Welcome to the HOW-TO video for Problem No. 2 by Cite PREP GUIDES for Math.

\LARGE \limn \ \frac{n^3}{e^{n/2}}

\LARGE \limn \ \frac{n^3}{e^{n/2}}

\LARGE \begin{gather*} a_n = \frac{n^3}{e^{n/2}} \end{gather*}

\LARGE \begin{gather*} \textsf{Given } a_n = \frac{n^3}{e^{n/2}} \ ,\\[1em] \textsf{find } \limn a_n \ . \end{gather*}

In this extended video, I consider three questions motivated by the How-To video for problem 2.

Extended How

Problem No. 2

Why is switching to a continuous function important?
Why does a quotient form imply L'Hopital's Rule?
Why do we repeat L'Hopital's Rule more than once?

Problem No. 2

Problem No. 2 involves finding the limit of the $\nth$ term of a sequence when that term is a quotient of two functions.

Let's get started.

Problem No. 2

When $a_n$ is a quotient

of two functions

Let's get started.

Solution

AAA For this problem, I see that $a_n$ is a quotient equal to a continuous function $f(x)$ for all positive $n,$

Quotient

\LARGE \begin{gather*} \textrm{from} \ \ a_n=\frac{n^3}{e^{n/2}}, \\[2em] \textrm{we have} \ \ a_n = f(n), \\[1em] f(x) = \frac{x^3}{e^{x/2}} , \ \ \ n \gt 0 \ . \end{gather*}

Why is this observation important?

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Given

Find

\LARGE \begin{gather*} a_n = \frac{n^3}{e^{n/2}} \\ f(x) = \frac{x^3}{e^{x/2}} \\ \end{gather*}

Quotient

\LARGE \begin{gather*} a_n = f(n), \\[1em] f(x) = \frac{x^3}{e^{x/2}} , \ \ \ n \gt 0 \ . \end{gather*}

Quotient

\LARGE \begin{gather*} a_n = f(n), \\[1em] f(x) = \frac{x^3}{e^{x/2}} , \ \ \ n \gt 0 \ . \end{gather*}

When I say that $a_n$ is a quotient equal to a continuous function $f(x)$ for all positive $n,$ the question is why is my observation important and how does that observation help us solve the problem.

Why?

The answer to that question is simple. If we can transform the $\nth$ term of a sequence from a discrete function of $n$ to a continuous function of $x,$

\LARGE \frac{n^3}{e^{n/2}} \ \ \ \Rightarrow \ \frac{x^3}{e^{x/2}} \ ,

then we can transform our limit problem to the limit of a continuous function,

\Large \limn \ \frac{n^3}{e^{n/2}} \ \ \ \Rightarrow \ \ \ \limx \ \frac{x^3}{e^{x/2}} \ .

Now, we can use additional methods to find the limit of the $\nth$ term that are otherwise not available, including

The Squeeze Theorem,
L'Hopital's Rule,
The limit of a composite function,
Direct substitution,
Definition of infinite limits,
Limit laws for continuous functions, and
The limits of trigonometric functions.

Back to
Problem 2

Moving on

with Problem 2

And because $a_n$ is a quotient, I suspect I will find the limit of a sequence using L'Hopital's Rule,

L'Hopital's Rule

\Large \begin{array}{l l} \textrm{If} \\ & \lima \ \frac {f(x)}{g(x)} = \frac \infty\infty\ , \\[2em] \textrm{then} \\ & \lima \ \frac {f(x)}{g(x)} \ = \ \lima \ \frac {f'(x)}{g'(x)} \ . \end{array}

Why does a quotient imply L'Hopital's Rule?

When a limit problem involves a quotient of two functions, discrete or continuous, I anticipate finding the limit may involve using L'Hopital's Rule.

The question is why can I draw this conclusion about using L'Hopital's Rule simply from a quotient in the problem.

The answer to this question is simple. If a limit problem involves a quotient, I immediately think of L'Hopital's Rule because the definition of the rule begins with an indeterminant form of a quotient.

Why?

Recall that L'Hopital's Rule says that if the limit of a quotient by direct subsitution yields an indeterminant form,

Indeterminant form

\LARGE \begin{gather*} \lima \ \frac {f(x)}{g(x)} \ = \ \frac \infty\infty \ , \end{gather*}

L'Hopital's Rule

\Large \begin{array}{l l} \textrm{If} \\ & \lima \ \frac {f(x)}{g(x)} = \frac \infty\infty\ , \\[2em] \textrm{then} \\ & \lima \ \frac {f(x)}{g(x)} \ = \ \lima \ \frac {f'(x)}{g'(x)} \ . \end{array}

then by L'Hopital's Rule you can take the derivative of the numerator and denominator individually, and then re-evaluate your limit by direct subsitution,

Take derivatives

\LARGE \lima \ \frac {f'(x)}{g'(x)} \ = \ L.

L'Hopital's Rule applies to other indeterminant forms aside from the quotient form above, including when

Other indeterminant forms

\LARGE \begin{gather*} \lima \ \frac {f(x)}{g(x)} \ = \ \frac {\pm \infty}{\pm \infty}, \\[2em] \lima \ \frac {f(x)}{g(x)} \ = \ \frac 00, \textrm{ and }\\[2em] \lima \ f(x) \cdot g(x) \ = \ 0 \cdot \infty\ . \end{gather*}

Steps to a Solution Z

Let $f(x)$ be a continuous function so that $a_n$ equals $f(x)$ for all positive values of $n.$

1. Let f(x)f(x) be a continuous function.
f(x)=x3ex/2,an=f(x),  n>0.\LARGE
\begin{gather*}
\LARGE f(x) = \frac {x^3}{e^{x/2}},\\[1em]
a_n = f(x), \ \ n \gt 0.
\end{gather*}

Apply L'Hopital's Rule as needed until the limit found is not indeterminant,

\Large \begin{gather*} \limx f(x) \ne \frac \infty\infty, \\[1em] \limx f(x) \ne \frac 00, \textrm{ or } \\[1em] \limx f(x) \ne 0 \cdot \infty. \end{gather*}

Why do we repeat L'Hopital's Rule?

Why?

For this problem, I say that we need to apply L'Hopital's Rule several times until we reach our answer. The question is why do we need to repeat L'Hopital's Rule several times for one problem, but not for others.

The answer to that question depends on the problem. Unfortunately, each time we apply L'Hopital's Rule, we end up with an indeterminant form, until we don't.

\LARGE \begin{gather*} \limx \frac{x^3}{e^{x/2}} \ = \ \frac \infty\infty \ ,\\[2em] \limx \frac{\ddx \ (x^3)}{\ddx \ (e^{x/2})} \ = \ \frac \infty\infty \ ,\\[2em] \limx \frac{\ddxb \ (x^3)}{\ddxb \ (e^{x/2})} \ = \ \frac \infty\infty \ ,\\[2em] \limx \frac{\ddxc \ (x^3)}{\ddxc \ (e^{x/2})} \ = \ \frac 6\infty \ = \ 0\ .\\[1em] \end{gather*}

For our problem, taking successive derivatives of the numerator reduces the degree of the numerator by one each time we take a derivative,

Numerator
Successive derivatives

\LARGE x^3 \ \Rightarrow \ 3x^2 \ \Rightarrow \ 6x \ \Rightarrow \ 6.

On the other hand, taking successive derivatives of the denominator changes the leading constant only,

Denominator
Successive derivatives

\LARGE \begin{gather*} e^{kx} \ \Rightarrow k^1e^{kx} \ \Rightarrow k^2e^{kx} \ \Rightarrow k^3e^{kx} \ \Rightarrow \tripledot \ , \\[1em] k= 1/2. \end{gather*}

Going forward, try to visualize successive applications of the derivative when applying L'Hopital's Rule for similar problems,

Visualize
successive derivatives

\LARGE \begin{gather*} \frac{x^3}{e^{kx}} \ \Rightarrow \ \frac{3x^2}{ke^{kx}} \ \Rightarrow \ \frac{6x}{k^2e^{kx}} \ \Rightarrow \ \frac{6}{k^3e^{kx}} \ ,\\[1em] \textrm{where } \limx \ \frac{6}{k^3e^{kx}} = 0 , \ k=1/2. \end{gather*}

We begin by finding the limit of $f(x)$ by direct substitution.

\Large \begin{gathered} \textrm{Begin by direct substitution} \\[0.5em] \begin{aligned} \limx \frac {x^3}{e^{x/2}} \end{aligned}\\[3em] \end{gathered}

As noted here, we find our result is an indeterminant form.

\Large \begin{gathered} \textrm{Begin by direct substitution} \\[0.5em] \begin{aligned} \limx \frac {x^3}{e^{x/2}} =\ \frac \infty\infty \end{aligned}\\[3em] \end{gathered}

So now we apply L'Hopital's Rule for the first time.

By L'Hopital's Rule, we know that if by direct substitution the limit of a function $f(x)$ takes an indeterminant form then the limit of $f(x)$ equals the limit of a new function defined as the derivative of the numerator divided by the derviative of the denominator.

\Large \begin{gathered} \textrm{Apply L'Hopital's Rule (first time)}\\[0.5em] \begin{aligned} \color{grey} \limx \frac {x^3}{e^{x/2}} =\ \frac \infty\infty \ \ \Rightarrow \ \ \\[1.5em] \color{black} \limx \frac {x^3}{e^{x/2}} =& \limn \frac{\ddx (x^3)}{\ddx (e^{x/2})}\\[2em] =& \limx \frac {3x^2}{(1/2)e^{x/2}} = \ \frac \infty\infty \end{aligned}\\[2em] \end{gathered}

Unfortunately, the result from applying L'Hopital's Rule the first time is yet another indeterminant form. Therefore, we must apply L'Hopital's Rule a second time.

As before, we apply L'Hopital's Rule by taking the derivative of the numerator and denominator independently, then finding a new limit by direct substitution.

\Large \begin{gathered} \textrm{Re-Apply L'Hopital's Rule (second time)} \\[0.5em] \begin{aligned} \color{grey} \limx \frac {3x^2}{(1/2)e^{x/2}} =\frac \infty\infty \ \ \Rightarrow \ \ \\[1.5em] \color{black} \limx \frac {3x^2}{(1/2)e^{x/2}} =& \limn \frac{\ddx (3x^2)}{\ddx (1/2)e^{x/2}}\\[2em] =& \limx \frac {6x}{(1/2)^2e^{x/2}} = \ \frac \infty\infty \end{aligned}\\[2em] \end{gathered}

Unfortunately again, the result of our second application of L'Hopital's Rule is yet another indeterminant form.

Applying L'Hopital's Rule a third time will do the trick. Upon our third application of L'Hopital's Rule our result is a limit of zero.

\Large \begin{gathered} \textrm{Re-Apply L'Hopital's Rule (third time)} \\[0.5em] \begin{aligned} \color{grey} \limx \frac {6x}{(1/2)^2e^{x/2}} = \ \frac \infty\infty \ \Rightarrow \ \\[1.5em] \color{black} \limx \frac {6x}{(1/2)^2e^{x/2}} =& \limn \frac{\ddx (6x)}{\ddx (1/2)^2e^{x/2}}\\[2em] =& \limx \frac {6}{(1/2)^3e^{x/2}} = 0 \end{aligned}\\[2em] \end{gathered}

2. Find lim⁡x→∞f(x)\limx f(x) by L'Hopital's Rule.
lim⁡x→∞f(x)=0\LARGE 
\limx f(x) = 0

Finally, by the Theorem for the Limit of a Sequence, we can say that the limit of $a_n$ equals the limit of $f(x)$ equals zero.

3. Find the Limit of a Sequence lim⁡n→∞an.\limn a_n.
lim⁡n→∞an= lim⁡x→∞f(x)lim⁡n→∞an= 0\LARGE
\begin{align*}
\LARGE 
\limn a_n =& \ \limx f(x) \\[1em]
\limn a_n =& \ 0
\end{align*}

\LARGE \begin{gather*} \limn a_n = 0 \end{gather*}