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Under which conditions

$$\lim_{a\to+\infty}\ln(f(a,x)) = \ln(z(x))\Longrightarrow \underset{a\to+\infty}{\lim}f(a,x) = z(x)\;?$$

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Was the switching of the parameters $a,x$ in $f$ in the second limit intentional? –  TMM Oct 21 '12 at 13:13
    
No, it should read as $f(a,x)$ or $f(x,a)$ in both cases. –  arkadiy Oct 21 '12 at 13:15
    
Put it simpler I just want to know if I find a limit of log of something is equal to the log. Can I infer that limit of the expressions under logs are equal. –  arkadiy Oct 21 '12 at 13:16
    
Or it would be simpler if the following is true $\lim_{a\to+\infty}\ln(f(a,x)) = \ln(\lim_{a\to+\infty} f(a,x))$. Is it always true and why? –  arkadiy Oct 21 '12 at 13:19

1 Answer 1

It's always true. You can take exponentials on both sides, and then pull the $\lim$ outside on the left-hand side since the exponential function is continuous.

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So can we claim that $$\lim_{a\to+\infty}g(f(a,x)) = g(z(x))\Longrightarrow \underset{a\to+\infty}{\lim}f(a,x) = z(x)\;?$$ for any continuous function $g(.)$? Which book should I consult to understand why this is so? –  arkadiy Oct 21 '12 at 18:28
    
@arkadiy: No, certainly not! But it's true for any invertible function with continuous inverse, since then you can apply $g^{-1}$ to both sides, etc. You don't really need to consult a book, since that's all there is to this argument, but any calculus textbook should do if you insist. ;-) –  Hans Lundmark Oct 21 '12 at 18:54
    
Now I got it. Thank you very much! No need for a calculus textbook.;) –  arkadiy Oct 21 '12 at 23:45

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