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Someone asks the following question: Determine the values for the real numbers $a$ and $b$ such that $$\lim_{x\to0}\frac{ae^x+b\cos x}{x}=2$$ The students directly apply de LHospital's rule for the quotient. I asked them to justify why by applying this rule is sufficient and i got no answer.

How would you explain this to a student?

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Remind the students of the requirements for L'Hôpital's rule. Use the requirements as stated in your textbook--the requirements may have been simplified.

The students should quickly see that one requirement is that the limit of the numerator of the fraction must be zero or infinity. The textbook may leave out infinity--if infinity is included, point out the limit of this particular numerator cannot be infinity. Then have the students find the limit of the numerator and set that to zero to get another requirement on $a$ and $b$.

In other words, go back to the definition (of L'Hôpital's rule), which is a major problem solving technique. You could emphasize the importance of checking conditions by showing the old "proof" that zero equal one (the proof uses division by zero).

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    $\begingroup$ how about this? My attempt was: the function $f(x)=x$ is differentiable and $f'(x)\neq0$ in every neighborhood of $0.$ Also $$\lim_{x\to0}f(x)=0$$ Now, define the function $g(x)=ae^x+b\cos x$. Then $$\lim_{x\to0}g(x)=0\Leftrightarrow a=-b$$ Then, IF we suppose that $a=-b,$ then we can apply DeLHospital's rule and show that $$\lim_{x\to0}\frac{ae^x-a\cos x}{x}=\lim_{x\to0}(ae^x+a\sin x)$$ Then $2=a$ and thus $b=-2.$ By the uniqueness of the limit, $2=a=-b.$ $\endgroup$
    – user12800
    Oct 2, 2019 at 1:06
  • $\begingroup$ @user12800: That looks fine. $\endgroup$
    – Rory Daulton
    Oct 2, 2019 at 7:49
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L'Hopital's rule is a horrid thing that should be banned from introductory calculus courses because it can and is used by those who do not understand it, and the problems it is used to solve are generally better solved by other means.

This question should be asked only of students who understand the Taylor approximations of the exponential and cosine. Then, since

\begin{align*} ae^{x} + b\cos{x} &= a( 1+ x + \text{terms quadratic in $x$}) + b(1 + \text{terms quadratic in $x$}) \\ & = a + ax + b + \text{terms quadratic in $x$} \end{align*} it follows that $$\frac{ae^{x} + b\cos{x}}{x} = \frac{a+b}{x} + a + \text{terms linear in $x$}$$ and so to obtain $2$ when $x \to 0$ it must be that $b = -a$ and $a = 2$.

Of course this is not a rigorous or careful argument, but all the understanding is there.

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    $\begingroup$ I understand what you are saying, but one may claim that since Taylor approximation 'requires' knowledge of antiderivatives and linear approximation, this may not be appropriate for students. But it is in the nature of Mathematics that nothing goes wasted! $\endgroup$
    – user12800
    Oct 2, 2019 at 12:52
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    $\begingroup$ L'Hopital requires those things too ... $\endgroup$
    – Dan Fox
    Oct 2, 2019 at 17:01
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Define $f(x):=ae^x+b\cos x,\,g(x)=x$. As $x\to0$, $\color{blue}{g\to0}$ while $\color{blue}{g^\prime\to1\ne0}$, and $\frac{f}{g}\to2$, so $\color{blue}{f\to2\times0=0}$. The blue conditions are what need to be checked for the rule's antecedent.

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