Using l'Hospitals rule for sequences of functions

[Whistlekins]

Homework Statement

Let (f_n) be the sequence of functions defined on [0,1] by f_n(x) = nxⁿlog(x) if x>0 and f_n(0)=0. Each f_n is continuous on [0,1].

Let a be in (0,1) and h_a(y) = ya^y := ye^{y log(a)}. Using l'Hospitals rule or otherwise, prove that lim_y->+∞ h_a(y) = 0.

Then considering different cases for x and using the previous part, prove that (f_n) converges point wise to the zero function.

Homework Equations

The Attempt at a Solution

I'm not too sure how to start this off. I rearranged h to a^y/(1/y), differentiated top and bottom so I get a^y/(-1/y²), and then taking the limit I get 0/0, and I'm not sure what to make of that. Also trying to rearrange the second part of h I get 0/0 when taking the limit of its derivatives.

What other path can I take to prove that the limit is 0?

 

[Dick]

Homework Statement

Let (f_n) be the sequence of functions defined on [0,1] by f_n(x) = nxⁿlog(x) if x>0 and f_n(0)=0. Each f_n is continuous on [0,1].

Let a be in (0,1) and h_a(y) = ya^y := ye^{y log(a)}. Using l'Hospitals rule or otherwise, prove that lim_y->+∞ h_a(y) = 0.

Then considering different cases for x and using the previous part, prove that (f_n) converges point wise to the zero function.

Homework Equations

The Attempt at a Solution

I'm not too sure how to start this off. I rearranged h to a^y/(1/y), differentiated top and bottom so I get a^y/(-1/y²), and then taking the limit I get 0/0, and I'm not sure what to make of that. Also trying to rearrange the second part of h I get 0/0 when taking the limit of its derivatives.

What other path can I take to prove that the limit is 0?

Take the log of y*a^y. Try to show it approaches -infinity as y->infinity. Use that log(a) is negative and show lim y->infinity log(y)/y=0.

 

[Whistlekins]

Take the log of y*a^y. Try to show it approaches -infinity as y->infinity. Use that log(a) is negative and show lim y->infinity log(y)/y=0.

I don't understand why I need to take the log. Aren't I changing the function then?

 

[Office_Shredder]

If you really want to use L'Hopital, write it as
\frac{y}{a^{-y}}

 

[Dick]

If you really want to use L'Hopital, write it as
\frac{y}{a^{-y}}

That is more economical. But my point for Whistlekins is that if log(r)->-infinity then r goes to 0.

 

[Whistlekins]

That is more economical. But my point for Whistlekins is that if log(r)->-infinity then r goes to 0.

Understood.

Now the second part of the question, can I just use the fact that limits of the product of two functions is equal to the product of the limits of the functions, and the fact that g_n(x) = nx^n is a subsequence of h? Is this all that is required for a proof?

 

[Dick]

Understood.

Now the second part of the question, can I just use the fact that limits of the product of two functions is equal to the product of the limits of the functions, and the fact that g_n(x) = nx^n is a subsequence of h? Is this all that is required for a proof?

Yes, I think so. Now you know nx^n->0 for x in (0,1). So nx^nlog(x) must do the same. x=0 and x=1 are separate cases.