Claim C: Assume the function f is differentiable and that lim f(x) = oo. f(x) lim exists lim f'(r) exists "Proof": We can use L'Hôpital's Rule: f(x) f'(x) lim lim lim 1 = lim f'(x) %3D %3D I00 Explain the error in the proof. Then prove that the claim is false with a counterexample.

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 $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{ }\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

Now, $\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{ }\mathbf{=}\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\infty }$ and $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{ }\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ is differentiable.

And, 

$$\begin{gathered} \mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\frac{\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{x}}\mathbf{ }\mathbf{=}\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\frac{\mathbf{sin}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{x}}\mathbf{=}\mathbf{0} \\ \mathit{A}\mathit{s}\mathbf{ }\mathit{w}\mathit{e}\mathbf{ }\mathit{k}\mathit{n}\mathit{o}\mathit{w}\mathbf{ }\mathbf{-}\mathbf{1}\mathbf{\le }\mathbf{sin}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\le }\mathbf{1}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathit{n} \\ \frac{\mathbf{-}\mathbf{1}}{\mathbf{x}}\mathbf{\le }\frac{\mathbf{sin}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{x}}\mathbf{\le }\frac{\mathbf{1}\mathbf{ }}{\mathbf{x}} \\ \mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\left(\frac{-1}{x}\right)\mathbf{ }\mathbf{\le }\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\frac{\mathbf{sin}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{x}}\mathbf{\le }\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\left(\frac{1}{x}\right) \\ \mathbf{\Rightarrow }\mathbf{0}\mathbf{\le }\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\frac{\mathbf{sin}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{x}}\mathbf{\le }\mathbf{0} \\ \mathbf{\text{by squeeze Theorem}}\mathbf{,} \\ \mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\frac{\mathbf{sin}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{x}}\mathbf{=}\mathbf{0} \end{gathered}$$

Here, we can see $\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\frac{\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{ }}{\mathbf{x}}$ exist.

Step:-2

$$\begin{gathered} As we know f\left(x\right)=\sin \left(x\right) \\ then f\text{'}\left(x\right)=\cos \left(x\right) \\ \mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{ }\mathbf{=}\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\mathbf{cos}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\infty } \end{gathered}$$

So, the claim is not true.