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Suppose $f(x)$ is differentiable at $x_0$ and $f(x_0)=1$. Find $\lim_{t\to0}f(x_0+t)^\frac{1}{t}$.

I tried using the epsilon delta definitions but I'm quite stuck.

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    $\begingroup$ Did you try using the definition of differentiable? In other words that there is a real number $f'(x_0)$ such that $f(x_0 + t)\approx f(x_0) + tf'(x_0)$ for small $t$? $\endgroup$ – Arthur Mar 9 '18 at 9:30
  • $\begingroup$ @Arthur that's also a good way! $\endgroup$ – user Mar 9 '18 at 10:07
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HINT

Note that

$$f(x_0+t)^\frac{1}{t} =e^{\frac{\log(f(x_0+t))}{t}}=e^{\frac{\log(f(x_0+t))-\log f(x_0)}{t-0}}$$

then note that exponential is continuos and evaluate

$$\lim_{t\to0} \frac{\log(f(x_0+t))-\log f(x_0)}{t-0} $$

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  • $\begingroup$ @Taroccoesbrocco this is an hint indeed, I don't want to give the full answer, the asker must work on it! $\endgroup$ – user Mar 9 '18 at 9:57
  • $\begingroup$ I see and it is a pertinent hint. In my opinion (maybe I'm wrong), it should be a comment, not an answer. Anyway, I haven't downvoted your hint. $\endgroup$ – Taroccoesbrocco Mar 9 '18 at 10:02
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    $\begingroup$ @Taroccoesbrocco A good hint is better than an answer when the asker is facing with a problem to solve and maybe with an homework. $\endgroup$ – user Mar 9 '18 at 10:05
  • $\begingroup$ @Taroccoesbrocco I've just added some detail, but no more than that. $\endgroup$ – user Mar 9 '18 at 10:11
  • $\begingroup$ I think now it is an answer that everybody can understand. Well done! I withdraw my negative review and upvote your answer. $\endgroup$ – Taroccoesbrocco Mar 9 '18 at 10:13

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