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Limit of an indeterminate expression

  • Thread starter ehrenfest
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  • #1
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Homework Statement


Let
[tex]U(t) = \frac{b^{t+1} - a^{t+1}}{(1+t)(b-a)}[/tex]

My book says that one can use "standard calculus methods to evaluate the limit of the indeterminate expression":

[tex]lim_{t\to0}U(t)^{1/t}[/tex]

Anyone want to fill me in on these methods? And what exactly is an indeterminate expression?


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
VietDao29
Homework Helper
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Homework Statement


Let
[tex]U(t) = \frac{b^{t+1} - a^{t+1}}{(1+t)(b-a)}[/tex]

My book says that one can use "standard calculus methods to evaluate the limit of the indeterminate expression":

[tex]lim_{t\to0}U(t)^{1/t}[/tex]

Anyone want to fill me in on these methods? And what exactly is an indeterminate expression?


Homework Equations





The Attempt at a Solution

Well, indeterminate expression is the one that you cannot plug the value in to find the limit. e.g, the expression [tex]\lim_{x \rightarrow 0} \frac{x ^ 3}{x + x ^ 2}[/tex] is an indeterminate one. If you plug x = 0 in, you'll get 0/0. That does not mean anything.

There are some Indeterminate Forms: [tex]\frac{0}{0} ; \ \frac{\infty}{\infty} ; 0 \times \infty ; \ \infty - \infty ; \mbox{and } 1 ^ \infty[/tex]. All of which have different ways to solve.

Your expression:
[tex]\lim_{t \rightarrow 0} \left( \frac{b ^ {t + 1} - a ^ {t + 1}}{(1 + t) (b - a)} \right) ^ {\frac{1}{t}}[/tex] is, in fact, the Indeterminate Form [tex]1 ^ \infty[/tex]. Can you see why?

Ok, to solve the Indeterminate Form [tex]1 ^ \infty[/tex], one should use the definition of e, the two definitions below are equivalent:

1. [tex]\lim_{x \rightarrow \infty} \left(1 + \frac{1}{x} \right) ^ x = e[/tex]

2. [tex]\lim_{x \rightarrow 0} \left(1 + x \right) ^ \frac{1}{x} = e[/tex]

e is, sometimes, defined as the 2 limits above, they both are in [tex]1 ^ \infty[/tex] form.

From the limit (2) above, we can easily prove that:
3. [tex]\lim_{x \rightarrow 0} \left(1 + \alpha x \right) ^ \frac{1}{x} = \lim_{x \rightarrow 0} \left(1 + \alpha x \right) ^ {\frac{1}{\alpha x} \times \alpha} \lim_{x \rightarrow 0} \left[ \left(1 + \alpha x \right) ^ {\frac{1}{\alpha x}} \right] ^ \alpha = e ^ \alpha[/tex]

-------------------------

Ok, now, back to our problem:
[tex]\lim_{t \rightarrow 0} \left( \frac{b ^ {t + 1} - a ^ {t + 1}}{(1 + t) (b - a)} \right) ^ {\frac{1}{t}}[/tex] can be splitted into 2 smaller limits, like this:

[tex]\lim_{t \rightarrow 0} \left( \frac{b ^ {t + 1} - a ^ {t + 1}}{(1 + t) (b - a)} \right) ^ {\frac{1}{t}} = \lim_{t \rightarrow 0} \left( \frac{1}{t + 1} \right) ^ \frac{1}{t} \times \lim_{t \rightarrow 0} \left( \frac{b ^ {t + 1} - a ^ {t + 1}}{b - a} \right) ^ \frac{1}{t}[/tex]

The first limit can be easily done by noticing that:

[tex]\lim_{t \rightarrow 0} \left( \frac{1}{1 + t} \right) ^ \frac{1}{t} = \lim_{t \rightarrow 0} \frac{1}{(1 + t) ^ \frac{1}{t}} = ?[/tex]. Can you finish it? :)





The second part is a little bit tricky, we must first convert it to polynomial before we can apply the limit (1), and (2). We know that the Taylor expansion for ex for x near 0 is:

[tex]e ^ x = 1 + x + \frac{x ^ 2}{2!} + \frac{x ^ 3}{3!} + ...[/tex]

Now, since [tex]b ^ t = e ^ {t \ln b}[/tex], and when t is near 0, t ln(b) is also near 0, so plus x = t ln(b), we'll have the expansion for bt for t near 0.

[tex]b ^ t = e ^ {t \ln b} = 1 + t \ln b + \frac{t ^ 2 \ln ^ 2 b}{2!} + \frac{t ^ 3 \ln ^ 3 t}{3!} + ...[/tex].

So we have: [tex]b ^ {t + 1} = b \times b ^ t = b \times \left( 1 + t \ln b + \frac{t ^ 2 \ln ^ 2 b}{2!} + \frac{t ^ 3 \ln ^ 3 t}{3!} + ... \right) = b + (b \ln b ) t + b \frac{\ln ^ 2 b}{2!} t ^ 2 + b \frac{\ln ^ 3 b}{3!}t ^ 2 + ...[/tex]. Now, we can truncate the series at anywhere, we'd like, and drop all the unnecessary terms (since the problem ask for the limit as t tends to 0).

Do the same to at + 1, we have:
[tex]a ^ {t + 1} = a + (a \ln a ) t + a \frac{\ln ^ 2 a}{2!} t ^ 2 + a \frac{\ln ^ 3 a}{3!}t ^ 2 + ...[/tex].

Now, we'll not find the limit of:
[tex]\lim_{t \rightarrow 0} \left(1 + \alpha t + \mathcal{O}(t ^ 2) \right) ^ \frac{1}{t}[/tex], where [tex]\alpha[/tex] is any constant, and [tex]\mathcal{O} (t)[/tex] is the big-oh notation, it satisfies: [tex]\lim_{t \rightarrow 0} \frac{\mathcal{O} (t ^ 2 )}{t} = 0[/tex]

[tex]\lim_{t \rightarrow 0} \left(1 + \alpha t + \mathcal{O}(t ^ 2) \right) ^ \frac{1}{t} = \lim_{t \rightarrow 0} \left(1 + \alpha t + \mathcal{O}(t ^ 2) \right) ^ {\frac{1}{\alpha t + \mathcal{O} (t ^ 2)} \times \frac{ \alpha t + \mathcal{O} (t ^ 2)}{t}}[/tex]

[tex]= \lim_{t \rightarrow 0} \left[ \left(1 + \alpha t + \mathcal{O}(t ^ 2) \right) ^ {\frac{1}{ \alpha t + \mathcal{O} (t ^ 2)}} \right] ^ \frac{\alpha t + \mathcal{O} (t ^ 2)}{t} = e ^ {\lim_{t \rightarrow 0} \frac{\alpha t + \mathcal{O} (t ^ 2 )}{t}} = e ^ \alpha = \lim (1 + \alpha t) ^ \frac{1}{t}[/tex]

So, the term [tex]\mathcal{O} (t ^ 2)[/tex] does not affect the final limits. So, we can drop all the terms, whose limit when divided by t, (and t tends to 0), tends to 0. So, we'll drop every terms that contain the power of t higher than 1. Can you see why?

----------------------------

[tex]\lim_{t \rightarrow 0} \left( \frac{b ^ {t + 1} - a ^ {t + 1}}{b - a} \right) ^ \frac{1}{t}[/tex]

[tex]= \lim_{t \rightarrow 0} \left( \frac{ b + (b \ln b) t - a - (a \ln a ) t + \mathcal{O} (t ^ 2)}{b - a} \right) ^ {\frac{1}{t}} = \lim_{t \rightarrow 0} \left(1 + \frac{ (b \ln b) t - (a \ln a ) t}{b - a} + \frac{\mathcal{O} (t ^ 2)}{b - a} \right) ^ {\frac{1}{t}}[/tex]

Drop the [tex]\mathcal{O} (t ^ 2)[/tex] part, we have:

[tex]= \lim_{t \rightarrow 0} \left(1 + \frac{ (b \ln b) t - (a \ln a ) t}{b - a} \right) ^ {\frac{1}{t}}[/tex]

Can you go from here? :)


----------------------------

Whoops, messing up with LaTeX. :cry: :cry: :cry:

Hopefully, you can get all of it, right? :)
 
Last edited:
  • #3
Gib Z
Homework Helper
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Very Nice work VietDao29 :) Though I would like to point out there was no need for the bold:

Taylor expansion for e^x for x near 0
The series converges throughout the real number line.
 
  • #4
VietDao29
Homework Helper
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...
The series converges throughout the real number line.
Well, yeah, I stand corrected. :blushing: :blushing:

I just wanna point out to the OP that, since the problem asks for the limit as t tends to 0, so he can drop all the terms he doesn't need, and keep the main ones. Well, I did un-bold them. :blushing:

Thanks.
 
  • #5
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Wow. Well done VietDao29. I am actually not entirely familiar with big O notation. I looked at wikipedia and that only confused me more. Can someone link me to a site that explains it better in the mathetmatical sense that VietDao29 is using it in and the computer science sense?
Thanks.
 
  • #6
VietDao29
Homework Helper
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Wow. Well done VietDao29. I am actually not entirely familiar with big O notation. I looked at wikipedia and that only confused me more. Can someone link me to a site that explains it better in the mathetmatical sense that VietDao29 is using it in and the computer science sense?
Thanks.
Well, Big Oh notation is a pretty hard concept to grasp. Wikipedia did explain it pretty well, but, it's pretty long, maybe that's what confuse you. If you have time, you can re-read the whole thing again, carefully, and I think you can get it.

So, if you haven't covered the Big Oh notation thingy yet, well, you can replace the Big Oh notation in the work above by some function h(t), which satisfies: [tex]\lim_{t \rightarrow 0} \frac{h(t)}{t} = 0[/tex], that's still work.

Cheers, :)
 

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