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If the limit of $f(x)$ exists and is finite, and the limit of $f'(x)$ exsits and is equal to $b$ then $b=0$

My Answer
Assume that $$\lim_{x \to \infty}f(x)=L$$ then $$\lim_{x \to \infty}\dfrac{f(x)-L}{x}=0$$ Also nituce that by L'Hopitals Rule $$\lim_{x \to \infty}\dfrac{f(x)-L}{x}=\lim_{x \to \infty}f'(x)=0$$ and so the above statement is true.

Is this proof correct??
Any feedback would be much appreciated.

otupygak
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  • Hint: use $\epsilon$-definition of the limit – afedder May 30 '14 at 21:19
  • Your solution is very good. If you want to avoid L'Hospital then you can argue as follows. We have $f(x) - f(x/2) = (x/2)f'(c)$ for some $c \in (x/2, x)$. Taking limits when $x \to \infty$ we see that LHS is $0$ and RHS is like $bx/2$ and hence $b$ must be $0$ (here $b = \lim_{x \to \infty}f'(x)$). – Paramanand Singh May 31 '14 at 04:28

1 Answers1

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It's fine, and is a good application of the version of L'Hôpital's rule mentioned in your previous question.

Just a nitpick: in your last display, it seems more natural to write $$b=\lim\limits_{x\rightarrow\infty} f'(x)=\lim\limits_{x\rightarrow\infty}{f(x)-L\over x}=0.$$

However, since many are unaware of the version of L'Hôpital mentioned in the link above, a better proof perhaps might be based on the Mean Value Theorem (applied to $f$ on intervals of the form $[N,N+1]$).

Community
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David Mitra
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