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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
$$ \lim_{x \to 0} f'(x) = \lim_{x \to 0} ( - \cos \frac{1}{x} + 2x \sin \frac{1}{x} )$$does not exist (because $$\lim_{x \to o} \cos \frac{1}{x}$$ 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 $$f(x)$$ and the limit of $$g(x)$$ both exist when $$x \rightarrow x_0$$ and then states that
$$\lim_{x \to x_0} ( f(x) + g(x) )$$
$$= \lim_{x \to x_0} f(x) + \lim_{x \to x_0} g(x)$$
$$= A + B $$
The way W&S use this theorem in Ex 4.5 above seems to be to say that if the
$$\lim_{x \to x_0} f(x)$$
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:
$$\lim_{x \to 0} x \sin \frac{1}{x} = 0
$$
If we took
$$f(x) = x$$ and $$g(x) = \sin \frac{1}{x}$$
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
$$\lim_{x \to 0} g(x) $$
$$= \lim_{x \to 0} \sin \frac{1}{x}
$$
does not exist, then neither does the limit of the product
$$\lim_{x \to 0} f(x) g(x)$$
$$= \lim_{x \to 0} x \sin \frac{1}{x}
$$
But we know that
$$\lim_{x \to 0} x \sin \frac{1}{x} = 0$$ !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
$$ \lim_{x \to 0} f'(x) = \lim_{x \to 0} ( - \cos \frac{1}{x} + 2x \sin \frac{1}{x} )$$does not exist (because $$\lim_{x \to o} \cos \frac{1}{x}$$ 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 $$f(x)$$ and the limit of $$g(x)$$ both exist when $$x \rightarrow x_0$$ and then states that
$$\lim_{x \to x_0} ( f(x) + g(x) )$$
$$= \lim_{x \to x_0} f(x) + \lim_{x \to x_0} g(x)$$
$$= A + B $$
The way W&S use this theorem in Ex 4.5 above seems to be to say that if the
$$\lim_{x \to x_0} f(x)$$
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:
$$\lim_{x \to 0} x \sin \frac{1}{x} = 0
$$
If we took
$$f(x) = x$$ and $$g(x) = \sin \frac{1}{x}$$
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
$$\lim_{x \to 0} g(x) $$
$$= \lim_{x \to 0} \sin \frac{1}{x}
$$
does not exist, then neither does the limit of the product
$$\lim_{x \to 0} f(x) g(x)$$
$$= \lim_{x \to 0} x \sin \frac{1}{x}
$$
But we know that
$$\lim_{x \to 0} x \sin \frac{1}{x} = 0$$ !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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