(c) Here is one more false claim and a bad proof.Claim C: ASsume the function f is differentiable and that lim f(x) = 00.f (x)limexistslim f'(x) exists"Proof": We can use L'Hôpital's Rule:f (x)lim(x)limf'(x)da= lim f'(x)1lim%3DdaExplain the error in the proof.Then prove that the claim is false with a counterexample.

Answered: Claim C: Assume the function f is…

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

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(c) Here is one more false claim and a bad proof.
Claim C: ASsume the function f is differentiable and that lim f(x) = 00.
f (x)
lim
exists
lim f'(x) exists
"Proof": We can use L'Hôpital's Rule:
f (x)
lim
(x)
lim
f'(x)
da
= lim f'(x)
1
lim
%3D
da
Explain the error in the proof.
Then prove that the claim is false with a counterexample.

Expert Solution

Step 1

Note:- Before explaining the error in proof i'll show you counter example. it'll help you to under stand the reason.

Step:-1

Counter Example:- Take $f (x) = s i n (x)$

Now, $l i m_{x \to \infty} f (x) = l i m_{x \to \infty} s i n (x) = \infty$ and $f (x) = s i n (x)$ is differentiable.

And,

$l i m_{x \to \infty} \frac{f (x)}{x} = l i m_{x \to \infty} \frac{\sin (x)}{x} = 0 A s w e k n o w - 1 \leq \sin (x) \leq 1 t h e n \frac{- 1}{x} \leq \frac{\sin (x)}{x} \leq \frac{1}{x} l i m_{x \to \infty} (\frac{- 1}{x}) \leq l i m_{x \to \infty} \frac{\sin (x)}{x} \leq l i m_{x \to \infty} (\frac{1}{x}) \Rightarrow 0 \leq l i m_{x \to \infty} \frac{\sin (x)}{x} \leq 0 by squeeze Theorem, l i m_{x \to \infty} \frac{\sin (x)}{x} = 0$

Here, we can see $l i m_{x \to \infty} \frac{f (x)}{x}$ exist.

Step:-2

$A s w e k n o w f (x) = \sin (x) t h e n f' (x) = \cos (x) l i m_{x \to \infty} f' (x) = l i m_{x \to \infty} \cos (x) = \infty$

So, the claim is not true.

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ISBN:

9780470458365

Author:

Erwin Kreyszig

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Wiley, John & Sons, Incorporated