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Im trying to solve some limits. This part of calculus is my weakest point, I simply do not have any idea how to solve this limit.

$$\lim_{x \to \infty} \frac{1-x}{2\sqrt\pi}\exp \left(-{\left({\frac{\log (x)-a}{b}}\right)}^2\right)$$

I do welcome any advices or solutions. Thanks for help!

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  • $\begingroup$ Is this concluded from normal distribution.? $\endgroup$
    – Nosrati
    Jan 23, 2017 at 11:23
  • $\begingroup$ It seems that the limit is $0$. $\endgroup$
    – Crostul
    Jan 23, 2017 at 12:30
  • $\begingroup$ No, its not from normal distribtion, but part of limit of another distribuiton. $\endgroup$
    – Bobesh
    Jan 23, 2017 at 12:52
  • $\begingroup$ Can you sow me why the limit is 0? $\endgroup$
    – Bobesh
    Jan 23, 2017 at 12:53

3 Answers 3

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Hint:

Taking the logarithm and ignoring the inessential constants, you get

$$\log x-\frac{\log^2x}{b^2}$$ which tends to $-\infty$ like

$$-\frac{\log^2x}{b^2}.$$

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It is not restrictive to assume $b>0$.

Do the substitution $(\log x-a)/b=t$; if $x\to\infty$, then also $t\to\infty$. Then $\log x=bt+a$, so we have $x=\exp(bt+a)$ and the limit becomes (leaving out the constant factor) $$ \lim_{t\to\infty}\frac{1-\exp(bt+a)}{\exp(t^2)} $$ With one application of l’Hôpital we get $$ \lim_{t\to\infty}\frac{-b}{2t\exp(t^2-bt-a)}=0 $$

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Hint. Consider the exponential law $\exp(x + y) = \exp(x)\exp(y)$ and that $\exp(\log(x)) = x$.

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