L’Hopital’s Limit From A Phone Game

One of my recently graduated students was playing a phone riddle game that required an answer to this level:

Turns out the solution to the game was just simply the word “answer” as the instruction was to simply enter that in. However, the integral and the limit shown there is possible to do and that is what this post is going to focus on. I don’t know what the game is called so I can’t link it. If I find out, I’ll edit this and link it in.

So here’s the problem in case you can’t see the screenshot:

Question

\[\lim_{x\rightarrow \infty}\frac{\int_1^x \left[t^2\left(e^\frac{1}{t}-1\right)-t\right]\;dt}{x^2\ln\left(1+\frac{1}{x}\right)}\]

Solution

Spoiler

The first thing to notice is that \(\displaystyle\lim_{x\rightarrow \infty} \left(1+\frac{1}{x}\right)^x\) is actually the definition of the constant \(e\).

By using logarithm laws, we can manipulate the denominator to become just \(x\):

\[\begin{align*}&\lim_{x\rightarrow \infty}\frac{\int_1^x \left[t^2\left(e^\frac{1}{t}-1\right)-t\right]\;dt}{x\ln\left(1+\frac{1}{x}\right)^x}\\=&\lim_{x\rightarrow \infty}\frac{\int_1^x \left[t^2\left(e^\frac{1}{t}-1\right)-t\right]\;dt}{x\ln\left(e\right)}\\=&\lim_{x\rightarrow \infty}\frac{\int_1^x \left[t^2\left(e^\frac{1}{t}-1\right)-t\right]\;dt}{x}\end{align*}\]

Now this is a \(\displaystyle \frac{\infty}{\infty}\) limit so we can use L’Hopital’s rule to help us evaluate the limit. If we differentiate the numerator, we apply the Fundamental Theorem of Calculus and the integral sign just goes away:

\[\begin{align*}&\lim_{x\rightarrow \infty}\frac{\left[x^2\left(e^\frac{1}{x}-1\right)-x\right]}{1}\\=&\lim_{x\rightarrow \infty}x^2\left(e^\frac{1}{x}-1\right)-x\\=&\lim_{x\rightarrow \infty}x^2e^\frac{1}{x}-x^2-x\end{align*}\]

We can algebraically manipulate this into a fraction:

\[\begin{align*}&\lim_{x\rightarrow \infty}\frac{e^\frac{1}{x}-1-\frac{1}{x}}{\frac{1}{x^2}}\end{align*}\]

This is a \(\displaystyle\frac{0}{0}\) limit so we can apply L’Hopital’s Rule:

\[\lim_{x\rightarrow \infty} \frac{\frac{-1}{x^2}e^{\frac{1}{x}}+\frac{1}{x^2}}{\frac{-2}{x^3}}\]

Simplifying this and then using L’Hopital’s Rule once more:

\[\begin{align*}&\lim_{x\rightarrow \infty} \frac{-e^{\frac{1}{x}}+1}{\frac{-2}{x}}\\=&\lim_{x\rightarrow \infty}\frac{\frac{1}{x^2}e^{\frac{1}{x}}}{\frac{2}{x^2}}\end{align*}\]

Finally, we can see that the factors of \(\displaystyle\frac{1}{x^2}\) cancel out leaving us with:

\[\lim_{x\rightarrow \infty} \frac{e^\frac{1}{x}}{2}=\frac{1}{2}\]

A way to check this answer is to just chuck it into a graphing application and check the horizontal asymptote:

Yep. The answer is definitely \(\frac{1}{2}\).

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