MHB Simple Exercise Involving the Algebra of Limits

[Math Amateur]

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:
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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:
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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

 

[Euge]

Hi Peter,

If the limit $\lim_{x\to 0} f(x)$ exists, then since $\lim_{x\to 0} 2x\sin \frac{1}{x}$ exists, Rule #2 implies $\lim_{x\to 0} \cos \frac{1}{x}$ exists, a contradiction. So, through indirect reasoning, we can see why the author made that comment (note: you were trying to use direct reasoning).

 

[Math Amateur]

Hi Peter,

If the limit $\lim_{x\to 0} f(x)$ exists, then since $\lim_{x\to 0} 2x\sin \frac{1}{x}$ exists, Rule #2 implies $\lim_{x\to 0} \cos \frac{1}{x}$ exists, a contradiction. So, through indirect reasoning, we can see why the author made that comment (note: you were trying to use direct reasoning).

Thanks Euge ...

Refelecting on this ... Trying to follow your post ... but having some trouble so far ...

Peter

 

[Euge]

Peter, please let me know where you're having trouble so I can assist you.

 

[Math Amateur]

Peter, please let me know where you're having trouble so I can assist you.

Thanks so much for your concern, Euge ... ...

However, after more reflection, I now follow your reasoning (which was pretty plausible anyway ... )

I was initially worried that what I thought was similar reasoning in the case of

$$\lim_{x \to 0} x \sin \frac{1}{x} = 0 $$

led to a wrong conclusion ... but I now see that this case is different ... the logic I was using was flawed and not the same as you used ... ...

Thanks again for your concern ... I really appreciate your help and support ...

Peter