Proof involving limit of derivative

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SUMMARY

The discussion focuses on identifying a function \( f \) for which the limit \( \lim_{x \to \infty} f(x) \) exists, while \( \lim_{x \to \infty} f'(x) \) does not. An example provided is \( f(x) = \sin(x) \), which oscillates between -1 and 1, thus converging in limit but having a derivative \( f'(x) = \cos(x) \) that does not converge. The discussion emphasizes the importance of using the chain rule in derivative calculations.

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Homework Statement



give an example of a function f for which lim f(x) as x[tex]\rightarrow[/tex][tex]\infty[/tex] exists, but lim f'(x) as x[tex]\rightarrow[/tex][tex]\infty[/tex] does not exist.

Homework Equations



f'(x) = lim [f(x+h)-f(x)]/h as h[tex]\rightarrow[/tex]0

The Attempt at a Solution



for some reason, i can only seem to find equations where lim f(x) as x[tex]\rightarrow[/tex][tex]\infty[/tex] does not exist, but lim f'(x) as x[tex]\rightarrow[/tex][tex]\infty[/tex] does exist. ex.f(x)=x and f'(x)=1
 
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Try thinking of functions that require the chain rule to find the derivative.
i.e. Let [itex]f(x) = g(h(x))[/itex]. Then [itex]f'(x) = g'(h(x)) \cdot h'(x)[/itex].
 

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