Calc Challenge IV: Evaluate Limit of Int.

[anemone]

Evaluate \(\displaystyle \lim_{{k}\to{\infty}} \int_{k}^{2k} \frac{k^3x}{x^5+1}\,dx\).

 

[MarkFL]

My solution:

Spoiler

I would write the limit as:

\(\displaystyle L=\lim_{k\to\infty}\left[\frac{\int_{k}^{2k}\dfrac{x}{x^5+1}}{k^{-3}}\right]\)

Observing that we have the indeterminate form 0/0, application of L'Hôpital's rule yields:

\(\displaystyle L=\lim_{k\to\infty}\left[\frac{\dfrac{k}{k^5+1}-\dfrac{4k}{32k^2+1}}{3k^{-4}}\right]\)

This can be simplified to:

\(\displaystyle L=\frac{1}{3}\lim_{k\to\infty}\left[\frac{k^5\left(28k^5-3\right)}{\left(32k^5+1\right)\left(k^5+1\right)}\right]=\frac{1}{3}\cdot\frac{28}{32}=\frac{7}{24}\)

 

[anemone]

My solution:

Spoiler

I would write the limit as:

\(\displaystyle L=\lim_{k\to\infty}\left[\frac{\int_{k}^{2k}\dfrac{x}{x^5+1}}{k^{-3}}\right]\)

Observing that we have the indeterminate form 0/0, application of L'Hôpital's rule yields:

\(\displaystyle L=\lim_{k\to\infty}\left[\frac{\dfrac{k}{k^5+1}-\dfrac{4k}{32k^2+1}}{3k^{-4}}\right]\)

This can be simplified to:

\(\displaystyle L=\frac{1}{3}\lim_{k\to\infty}\left[\frac{k^5\left(28k^5-3\right)}{\left(32k^5+1\right)\left(k^5+1\right)}\right]=\frac{1}{3}\cdot\frac{28}{32}=\frac{7}{24}\)

Aww...that is fast and efficient of you, MarkFL to solve my today challenge! Well done MarkFL!

When you solved it using the L'Hôpital's rule, your solution becomes very neat and straightforward. I am wondering if you want to tackle it for another time without using the help from the L'Hôpital's rule, hehehe...

 

[MarkFL]

Aww...that is fast and efficient of you, MarkFL to solve my today challenge! Well done MarkFL!

When you solved it using the L'Hôpital's rule, your solution becomes very neat and straightforward. I am wondering if you want to tackle it for another time without using the help from the L'Hôpital's rule, hehehe...

Why do you want to tie my hands like that? (Wasntme)

 

[anemone]

Why do you want to tie my hands like that? (Wasntme)

Because I would sing a song for you if you try that!:p

 

[anemone]

Solution without the use of L'Hôpital's rule:

Spoiler

Since $x^5+1<x(x+1)^4$ and $x^5<x^5+1$ for $x\ge 1$, we have

$\dfrac{1}{(x+1)^4}<\dfrac{x}{x^5+1}<\dfrac{1}{x^4}$ for $x\ge 1$

It then follows that

$\displaystyle k^3\int_{k}^{2k} \frac{1}{(x+1)^4}\,dx<k^3\int_{k}^{2k} \frac{x}{x^5+1}\,dx<k^3\int_{k}^{2k} \frac{1}{x^4}\,dx$ for every $k\ge 1$.

Now,

$\displaystyle \begin{align*} k^3\int_{k}^{2k} \frac{1}{(x+1)^4}\,dx&=\dfrac{k^3}{3}\left(\dfrac{1}{(k+1)^3}-\dfrac{1}{(2k+1)^3}\right)\\&=\dfrac{1}{3}\left(\left(\dfrac{k}{(k+1)}\right)^3-\left(\dfrac{k}{(2k+1)^3}\right)^3\right)\\&=\dfrac{1}{3}\left(1-\dfrac{1}{8}\right)\text{when $k$ approaches $\infty$}\\&=\dfrac{7}{24}\end{align*}$

and

$\displaystyle \begin{align*} k^3\int_{k}^{2k} \frac{1}{x^4}\,dx&=\dfrac{k^3}{3}\left(\dfrac{1}{k^3}-\dfrac{1}{(2k)^3}\right)\\&=\dfrac{1}{3}\left(1-\dfrac{1}{8}\right)\text{when $k$ approaches $\infty$}\\&=\dfrac{7}{24}\end{align*}$

It follows by the Squeeze principle that hence \(\displaystyle \lim_{{k}\to{\infty}} \int_{k}^{2k} \frac{k^3x}{x^5+1}\,dx=\dfrac{7}{24}\).