How about these? EXAMPLE 1: x/(h-x^2) His method gives 1/(h' - 2x) The correct answer is -1/x. His answer is incorrect even if you substitute h'=0. EXAMPLE 2: Or how about cos(h)/h ? The limit "is" infinity, but his method results in -sin(h) => -sin(0) = 0. EXAMPLE 3: Or how about a slight variation? cos(h)/(h+x) The limit is 1/x, but his method gives sin(h)*h' / (1+h') => 0. EXAMPLE 4: sin(x+h)/cos(x+h) Actual: tan(x) His: -cot(x) Are these sufficient? Robert Smith On 12/5/2011 12:43 PM, James Propp wrote:
A student in my honors calculus class just showed me his solutions to two homework problems I assigned; in each case he got the right answer for the wrong reason. I thought I could improvise an example to show him why his method (aside from including steps with no logical justification) would give the wrong answer if he tried applying it to other problems, but I failed, and I'm wondering if someone can provide such an example, or else explain why no such example exists.
The first homework problem was to evaluate the limit of (f(x+h)-f(x-h))/2h as h approaches 0. The student differentiated on top and bottom with respect to x (rather than h), obtaining lim (f'(x+h)(1+h')-f'(x-h)(1-h'))/2h' where h'=dh/dx. Substituting h=0 inside the limit gives (f'(x)(1+h')-f'(x)(1-h'))/2h' which simplifies to f'(x).
The second homework problem was to evaluate the limit of (f(x+h)-2f(x)+f(x-h))/h^2 as h approaches 0. The student again differentiated on top and bottom with respect to x (rather than h), obtaining lim (f'(x+h)(1+h') - 2f'(x) + f'(x-h)(1-h'))/2hh' The student then "applied L'Hospital" again to rewrite the limit as lim (f"(x+h)(1+h')(1+h') + f'(x+h)(h") - 2f"(x) + f"(x-h)(1-h')(1-h') + f'(x-h)(-h"))/(2hh"+2h'h'). Substituting h=0 inside the limit gives (f"(x)(1+h')(1+h') + f'(x)(h") - 2f"(x) + f"(x)(1-h')(1-h') + f'(x)(-h"))/(2h'h') which simplifies to f"(x).
The student half-understood my explanation of why both derivations are wrong, but I don't think he'll fully accept that his method is wrong until I show him an example where it leads to a wrong answer.
I just sent him an email, asking him to use his method to compute the limit of (x+2h)/(x+h) as h approaches 0. I suspect that he'll either get 1 or 1/2 as his answer, as opposed to the correct answer, which is "1 if x is non-zero, 2 if x is zero." But then he'll still feel that his approach has merit, because it gives the right answer for at least some values of x.
Is there a better example that shows that taking derivatives with respect to the wrong variable can give really, really wrong answers?
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