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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!

  • 0
    Is this concluded from normal distribution.? – 2017-01-23
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    It seems that the limit is $0$. – 2017-01-23
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    No, its not from normal distribtion, but part of limit of another distribuiton. – 2017-01-23
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    Can you sow me why the limit is 0? – 2017-01-23

3 Answers 3

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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 $$ \lim_{t\to\infty}\frac{-b}{2t\exp(t^2-bt-a)}=0 $$

1

Hint:

Taking the logarithm and ignoring the inessential constants, you get

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

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

0

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