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Is $\displaystyle \lim_{x \to a} \ln f(x)= \ln\lim_{x \to a}f(x)$ ?

Is this correct ? Are there any conditions that $f(x)$ must satisfy ?

Angelo Mark
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    A more general version of the question is: Is $\displaystyle \lim_{x \to a} g( f(x))=g(\lim_{x \to a}f(x))$

    A proof of this theorem may be found at http://math.stackexchange.com/questions/285667/how-to-prove-this-limit-composition-theorem, and an important cautionary counterexample is at http://math.stackexchange.com/questions/173781/is-there-a-better-counter-example-problem-involving-limit-of-composition-of-fu.

     – mweiss Nov 09 '15 at 19:29

2 Answers2

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This is always true (assuming either limit exists) because of the continuity and injectivity of the function $\ln(x)$. If we replaced $\ln$ with a non-injective, continuous function, the limit on the left might exist when the limit on the right doesn't.

Because $\ln$ is continuous, the limits exist if $\lim_{x \to a}f(x)$ exists and is positive. Because $\ln$ is injective, the converse holds as well.

Ben Grossmann
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  • You need the existence of bot limits or additionally injectivity of $\ln$ – Hagen von Eitzen Nov 09 '15 at 19:12
  • @HagenvonEitzen good point, added that in. – Ben Grossmann Nov 09 '15 at 19:16
  • I think being injective is not enough for the converse! I think the inverse function must be continuous so that we can use this function on the right side and move the inverse inside the limit! Is there another way that does not need the inverse to be continuous? – Dandelion Nov 27 '18 at 16:30
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If the limit of $f(x)$ exists and is greater than $0$, then yes. Otherwise no.

5xum
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