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JA7. An interesting problem

Statement

Choose one interesting problem from the text of medium difficulty that was not assigned.

  • Describe why you find it interesting.
  • Either solve it or find a solution online
  • Work through it using your own understanding to critique that solution and improve it.

Solution

Describe why you find it interesting

The problem number 3 in the written assignment was interesting, here is the problem statement (UoPeople, 2023):

3. Calculate the indicated limit. \(\displaystyle \lim_{x \to 0} \frac{sinx-x}{x^2}\) . If a limit does not exist then answer \(+ \infty , – \infty , or DNE\) (whichever is correct). Make sure to check that L’Hopital’s rule applies before using it. Also, whenever you apply L’Hopitals rule, indicate that you are using it.

The above problem seems simple at the beginning, just a straightforward application of L’Hopital’s rule, but it is not, as we will see.

Either solve it or find a solution online

let’s find the limit as x approaches 0:

\[ \begin{aligned} \lim_{x \to 0} \frac{sinx-x}{x^2} &= \frac{sin(0)-0}{0^2} \\ &= \frac{0}{0} \\ \end{aligned} \]

So the limit is 0/0, which is indeterminate. We can use L’Hopital’s rule:

\[ \begin{aligned} \lim_{x \to 0} \frac{sinx-x}{x^2} &= \lim_{x \to 0}\frac{(sinx-x)'}{(x^2)'} \\ &= \lim_{x \to 0} \frac{cosx-1}{2x} \\ &= \frac{cos(0)-1}{2(0)} \\ &= \frac{1-1}{0} \\ &= \frac{0}{0} \\ \end{aligned} \]

So after applying L’Hopital’s rule, we still have an indeterminate form; and I thought that since L’Hopital’s rule did not give us a determinate result, then the limit does not exist, and we should report that no answer was found.

Unfortunately, this was not right; and as long as the limit is still indeterminate, we can use L’Hopital’s rule again on the derivative of the previous step till we get a determinate result (Dawkins, 2022).

So we can apply L’Hopital’s rule again on \(\frac{cosx-1}{2x}\):

\[ \begin{aligned} \lim_{x \to 0} \frac{cosx-1}{2x} &= \lim_{x \to 0}\frac{(cosx-1)'}{(2x)'} \\ &= \lim_{x \to 0} \frac{-sinx}{2} \\ &= \frac{-sin(0)}{2} \\ &= \frac{0}{2} \\ &= 0 \\ \end{aligned} \]

So the limit of the original function is 0.

Work through it using your own understanding to critique that solution and improve it

The path to the problem consists of three steps:

  1. Try to find the limit in a direct way first.
  2. If the limit is indeterminate, we can use L’Hopital’s rule.
  3. If the limit is still indeterminate, we can use L’Hopital’s rule again on the derivative of the previous step till we get a determinate result.

At the start I did read L’Hopital’s rule properly, so I did know the that the third step was possible so it was my mistake, but I think problem statements can be useful in giving some hints about the tricks in the problem; because I would have easily submitted it after the second step reporting that the limit does not exist.

The third step also trebles me, because there are definitely some cases where the limit may go interminably forever; at this point how do we know where to stop? I think this is a good question to ask for people who are really interested in the topic.

Conclusion

The problem was interesting, and I learned from it as it represents a good example of how to use L’Hopital’s rule. However, this is also a reminder that the student need to carefully read all required readings before moving to solve problems, and for instructors to give some hints about the tricks in the problem statements. And of course, the ultimate lesson, is that there is always more to the topic than what we know.

References