## anonymous 3 years ago Let a>0 (a). Show using L'Hospitals rule that: $\lim_{x \rightarrow 0} x ^{a}\ln(x)=0$ (b). By setting x=ln(t) in (a). show that: $\lim_{x \rightarrow -\infty} |x|^{a} e ^{x}=0$ I need help with (b). I don't get the question, how do I show that (b). is true by setting x=ln(t) in (a). it doesn't make any sense to me.

1. anonymous

the proposed substitution

2. anonymous

$x=\ln t$ means that $t=e^x$

3. anonymous

I've got that but how can I use it?

4. anonymous

as $x \rightarrow -\infty, t \rightarrow ?$

5. anonymous

Zero?

6. anonymous

e^-infinty = 0

7. anonymous

correct

8. anonymous

so we shift from

9. anonymous

$\lim_{x \rightarrow -\infty}$ to $\lim_{t \rightarrow 0^+}$

10. anonymous

in terms of t, $\left| x \right|^a = ?$

11. anonymous

$|\ln (t)|^{a}$ ?

12. anonymous

that's right. and what about $e^x$

13. anonymous

We could set it as just t, right?

14. anonymous

right

15. anonymous

$\lim_{t \rightarrow 0} |\ln(t)|^{a}t$

16. anonymous

Because t->0 shouldn't the whole expression be equal to 0 then?

17. anonymous

not necessarily

18. anonymous

what about $\left| \ln t \right|^a$

19. anonymous

$\ln(t) \rightarrow -\infty$ but since it's the abolute value its just $\infty$

20. anonymous

$\left| \ln t \right|^a \rightarrow +\infty$. you are correct

21. anonymous

the transformed problem is an indeterminate form $0*\infty$ which should be resolved either as $\frac{ 0 }{ 0 }$ or $\frac{ \infty }{ \infty }$

22. anonymous

Right, so I could tranform it to the right form and use L'Hospitals rule to get the answer correct

23. anonymous

Should I use the form $\frac{ t }{ |\ln(t)|^{-a} }$ and then go for L'Hospital?

24. anonymous

25. anonymous

Just as it was, is it possible to use L'hostial on multiplication form?

26. anonymous

lhopitals rule must be in quotient form

27. anonymous

Which form are you thinking about then?

28. anonymous

29. anonymous

$-\frac{ a|\ln(t)|^{-a-1} }{ t }$ Can that be the correct derivation? If so it should give us the right answer $\lim_{t \rightarrow 0}\frac{ 1 }{ -\frac{ a|\ln(t)|^{-a-1} }{ t } } = \frac{ 1 }{- \infty }=0$ Can this be right?

30. anonymous

we'll look at the 2nd quotient form $\frac{ \left| ?\ln t \right|^a }{ \frac{ 1 }{ t } }$, noting that a>0

31. anonymous

since t->0+, $\left| \ln t \right|=-\ln t$

32. anonymous

Right

33. anonymous

the indeterminate form is inf/inf. so we'll apply l'hop's rule once

34. anonymous

i hate getting disconnected in the middle of the equation

35. anonymous

$\frac{ \frac{ -a \ln ^{a-1}(t) }{t } }{\frac{ -1 }{ t ^{2} } }$

36. anonymous

right. when simplified becomes $\frac{ a(-\ln t)^{a-1} }{ 1/t }$

37. anonymous

here's the catch to the problem

38. anonymous

$(-\ln t)^a \rightarrow +\infty$ for as long as the exponent a > 0.

39. anonymous

So it's still $\frac{ \infty }{ \infty }$

40. anonymous

right. but sooner or later, the exponent becomes negative after successive applications of l'hopital's rule

41. anonymous

$\frac{ (a-1)a(-\ln(t))^{a-2} }{ 1/t }$

42. anonymous

after k applications of the rule, the indeterminate form is now $a(a-1)(a-2)...(a-k+1)\frac{ (-\ln t)^{a-k} }{ 1/t }$ where a-k is negative

43. anonymous

so that $(-\ln t)^{a-k} \rightarrow 0$ because a-k is negative

44. anonymous

the quotient form is now $a(a-1)(a-2)...(a-k+1)\frac{ 0 }{ \infty }$ and the problem is solved!

45. anonymous

Oh, that's clever, I get it

46. anonymous

Thank you so much for you're help I can't thank you enought this one has been on my mind for a couple of days now!