I'd like to know how to calculate the following:

$$ \lim_{x \to \infty} \frac{1}{x^2}\log\left(\left(1+(\exp(2x)-1)^2\right)\right)^2. $$

Unfortunately, I'm not even sure where to begin with calculating this expression, other than just asking Wolfram Alpha. Any help would be much appreciated.

  • 1
    $\begingroup$ Hmm perhaps a variable change could help? Something like $$ x = \frac{1}{2}\ln(t+1) $$ and then instead of $x \to \infty$ you can look at $t \to \infty$ ... I dunno, just a thought. $\endgroup$
    – Matti P.
    Commented Nov 8, 2022 at 13:50
  • 1
    $\begingroup$ @MattiP. Thanks! In fact, I think that works - you can then use L'Hopital's rule to figure out the limit of log(1+t^2)/log(1+t). Well spotted! $\endgroup$
    – Chris
    Commented Nov 8, 2022 at 14:02

1 Answer 1


For anyone interested, I used Matti P.'s suggestion and it worked.

Set $x=\frac{1}{2}\log(t+1)$, then you get

$$ \lim_{x \to \infty}x^{-2}\log(1+(\exp(2x)-1))^2=\lim_{t \to \infty}\frac{4\log(t^2+1)^2}{\log(t+1)^2}. $$

Then using L'Hopital's rule,

$$ \lim_{t \to \infty}\frac{4\log(t^2+1)^2}{\log(t+1)^2}=\lim_{t \to \infty}\left(\frac{2t(t+1)}{1+t^2}\right)^2=4^2=16. $$



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