## frx Group Title 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. one year ago one year ago

1. sirm3d Group Title

the proposed substitution

2. sirm3d Group Title

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

3. frx Group Title

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

4. sirm3d Group Title

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

5. frx Group Title

Zero?

6. frx Group Title

e^-infinty = 0

7. sirm3d Group Title

correct

8. sirm3d Group Title

so we shift from

9. sirm3d Group Title

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

10. sirm3d Group Title

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

11. frx Group Title

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

12. sirm3d Group Title

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

13. frx Group Title

We could set it as just t, right?

14. sirm3d Group Title

right

15. frx Group Title

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

16. frx Group Title

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

17. sirm3d Group Title

not necessarily

18. sirm3d Group Title

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

19. frx Group Title

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

20. sirm3d Group Title

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

21. sirm3d Group Title

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

22. frx Group Title

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

23. frx Group Title

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

24. sirm3d Group Title

how about the other form?

25. frx Group Title

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

26. sirm3d Group Title

lhopitals rule must be in quotient form

27. frx Group Title

Which form are you thinking about then?

28. sirm3d Group Title

i'm trying your quotient form

29. frx Group Title

$-\frac{ a|\ln(t)|^{-a-1} }{ t }$ 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 Group Title

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

31. sirm3d Group Title

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

32. frx Group Title

Right

33. sirm3d Group Title

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

34. sirm3d Group Title

i hate getting disconnected in the middle of the equation

35. frx Group Title

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

36. sirm3d Group Title

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

37. sirm3d Group Title

here's the catch to the problem

38. sirm3d Group Title

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

39. frx Group Title

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

40. sirm3d Group Title

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

41. frx Group Title

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

42. sirm3d Group Title

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 Group Title

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

44. sirm3d Group Title

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

45. frx Group Title

Oh, that's clever, I get it

46. frx Group Title

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!