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I need some help in understanding the reasoning and analysis in the solution to Exercise 4.5 in Robert C. Wrede and Murray Spiegel's (W&S) book: "Advanced Calculus" (Schaum's Outlines Series).
Exercise 4.5 in W&S reads as follows:
https://www.physicsforums.com/attachments/3918
View attachment 3919
The reasoning/analysis that bothers me is when Wrede and Spiegel (W&S) write:
" ... ... Since
\(\displaystyle \lim_{x \to 0} f'(x) = \lim_{x \to 0} ( - \cos \frac{1}{x} + 2x \sin \frac{1}{x} )\)does not exist (because \(\displaystyle \lim_{x \to o} \cos \frac{1}{x}\) does not exist), ... ... "
The reasoning (despite its apparent plausibility) worries me ...
The reason this bothers me is that the above reasoning seems to draw on a theorem concerning the algebra of limits in a mistaken way. The particular theorem concerned states that the limit of a sum of functions is the sum of the limits ... ... indeed the theorem on the algebra of limits in W&S reads as follows:
https://www.physicsforums.com/attachments/3920
https://www.physicsforums.com/attachments/3921
BUT ... ... note that the theorem regarding the sum of limits assumes that the limit of \(\displaystyle f(x)\) and the limit of \(\displaystyle g(x)\) both exist when \(\displaystyle x \rightarrow x_0\) and then states that
\(\displaystyle \lim_{x \to x_0} ( f(x) + g(x) )\)
\(\displaystyle = \lim_{x \to x_0} f(x) + \lim_{x \to x_0} g(x)\)
\(\displaystyle = A + B \)
The way W&S use this theorem in Ex 4.5 above seems to be to say that if the
\(\displaystyle \lim_{x \to x_0} f(x)\)
does not exist then the limit of the sum does not exist ... ... but what is the exact/rigorous justification for this? ... ... it does not seem to follow the theorem I have quoted ... ...
(Mind you, as I have indicated above, what they say sounds eminently reasonable ... intuitively speaking ... !?)One of the reasons for my caution regarding W&S's reasoning in Exercise 4.5 is that I began thinking of the known limit:
\(\displaystyle \lim_{x \to 0} x \sin \frac{1}{x} = 0
\)
If we took
\(\displaystyle f(x) = x\) and \(\displaystyle g(x) = \sin \frac{1}{x}\)
and then applied the same reasoning as W&S above (although now applied to the limit of a product rather than the limit of a sum) then we could, I think, argue that because
\(\displaystyle \lim_{x \to 0} g(x) \)
\(\displaystyle = \lim_{x \to 0} \sin \frac{1}{x}
\)
does not exist, then neither does the limit of the product
\(\displaystyle \lim_{x \to 0} f(x) g(x)\)
\(\displaystyle = \lim_{x \to 0} x \sin \frac{1}{x}
\)
But we know that
\(\displaystyle \lim_{x \to 0} x \sin \frac{1}{x} = 0\) !However, surely this is the same reasoning as W&S apply to Exercise 4.5? Isn't it?Can someone please clarify the above?
Peter
Exercise 4.5 in W&S reads as follows:
https://www.physicsforums.com/attachments/3918
View attachment 3919
The reasoning/analysis that bothers me is when Wrede and Spiegel (W&S) write:
" ... ... Since
\(\displaystyle \lim_{x \to 0} f'(x) = \lim_{x \to 0} ( - \cos \frac{1}{x} + 2x \sin \frac{1}{x} )\)does not exist (because \(\displaystyle \lim_{x \to o} \cos \frac{1}{x}\) does not exist), ... ... "
The reasoning (despite its apparent plausibility) worries me ...
The reason this bothers me is that the above reasoning seems to draw on a theorem concerning the algebra of limits in a mistaken way. The particular theorem concerned states that the limit of a sum of functions is the sum of the limits ... ... indeed the theorem on the algebra of limits in W&S reads as follows:
https://www.physicsforums.com/attachments/3920
https://www.physicsforums.com/attachments/3921
BUT ... ... note that the theorem regarding the sum of limits assumes that the limit of \(\displaystyle f(x)\) and the limit of \(\displaystyle g(x)\) both exist when \(\displaystyle x \rightarrow x_0\) and then states that
\(\displaystyle \lim_{x \to x_0} ( f(x) + g(x) )\)
\(\displaystyle = \lim_{x \to x_0} f(x) + \lim_{x \to x_0} g(x)\)
\(\displaystyle = A + B \)
The way W&S use this theorem in Ex 4.5 above seems to be to say that if the
\(\displaystyle \lim_{x \to x_0} f(x)\)
does not exist then the limit of the sum does not exist ... ... but what is the exact/rigorous justification for this? ... ... it does not seem to follow the theorem I have quoted ... ...
(Mind you, as I have indicated above, what they say sounds eminently reasonable ... intuitively speaking ... !?)One of the reasons for my caution regarding W&S's reasoning in Exercise 4.5 is that I began thinking of the known limit:
\(\displaystyle \lim_{x \to 0} x \sin \frac{1}{x} = 0
\)
If we took
\(\displaystyle f(x) = x\) and \(\displaystyle g(x) = \sin \frac{1}{x}\)
and then applied the same reasoning as W&S above (although now applied to the limit of a product rather than the limit of a sum) then we could, I think, argue that because
\(\displaystyle \lim_{x \to 0} g(x) \)
\(\displaystyle = \lim_{x \to 0} \sin \frac{1}{x}
\)
does not exist, then neither does the limit of the product
\(\displaystyle \lim_{x \to 0} f(x) g(x)\)
\(\displaystyle = \lim_{x \to 0} x \sin \frac{1}{x}
\)
But we know that
\(\displaystyle \lim_{x \to 0} x \sin \frac{1}{x} = 0\) !However, surely this is the same reasoning as W&S apply to Exercise 4.5? Isn't it?Can someone please clarify the above?
Peter
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