## frx 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. sirm3d

the proposed substitution

2. sirm3d

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

3. frx

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

4. sirm3d

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

5. frx

Zero?

6. frx

e^-infinty = 0

7. sirm3d

correct

8. sirm3d

so we shift from

9. sirm3d

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

10. sirm3d

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

11. frx

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

12. sirm3d

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

13. frx

We could set it as just t, right?

14. sirm3d

right

15. frx

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

16. frx

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

17. sirm3d

not necessarily

18. sirm3d

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

19. frx

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

20. sirm3d

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

21. sirm3d

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

22. frx

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

23. frx

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

24. sirm3d

25. frx

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

26. sirm3d

lhopitals rule must be in quotient form

27. frx

Which form are you thinking about then?

28. sirm3d

29. frx

$-\frac{ a|\ln(t)|^{-a-1} }{ t }$ 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. sirm3d

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

31. sirm3d

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

32. frx

Right

33. sirm3d

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

34. sirm3d

i hate getting disconnected in the middle of the equation

35. frx

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

36. sirm3d

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

37. sirm3d

here's the catch to the problem

38. sirm3d

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

39. frx

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

40. sirm3d

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

41. frx

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

42. sirm3d

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. sirm3d

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

44. sirm3d

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

45. frx

Oh, that's clever, I get it

46. frx

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!