ehrenfest said:
Homework Statement
Let
U(t) = \frac{b^{t+1} - a^{t+1}}{(1+t)(b-a)}
My book says that one can use "standard calculus methods to evaluate the limit of the indeterminate expression":
lim_{t\to0}U(t)^{1/t}
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 into find the limit. e.g, the expression \lim_{x \rightarrow 0} \frac{x ^ 3}{x + x ^ 2} 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: \frac{0}{0} ; \ \frac{\infty}{\infty} ; 0 \times \infty ; \ \infty - \infty ; \mbox{and } 1 ^ \infty. All of which have different ways to solve.
Your expression:
\lim_{t \rightarrow 0} \left( \frac{b ^ {t + 1} - a ^ {t + 1}}{(1 + t) (b - a)} \right) ^ {\frac{1}{t}} is, in fact, the Indeterminate Form 1 ^ \infty. Can you see why?
Ok, to solve the Indeterminate Form 1 ^ \infty, one should use the definition of e, the two definitions below are equivalent:
1. \lim_{x \rightarrow \infty} \left(1 + \frac{1}{x} \right) ^ x = e
2. \lim_{x \rightarrow 0} \left(1 + x \right) ^ \frac{1}{x} = e
e is, sometimes, defined as the 2 limits above, they
both are in 1 ^ \infty form.
From the limit
(2) above, we can easily prove that:
3. \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
-------------------------
Ok, now, back to our problem:
\lim_{t \rightarrow 0} \left( \frac{b ^ {t + 1} - a ^ {t + 1}}{(1 + t) (b - a)} \right) ^ {\frac{1}{t}} can be splitted into 2 smaller limits, like this:
\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}
The first limit can be easily done by noticing that:
\lim_{t \rightarrow 0} \left( \frac{1}{1 + t} \right) ^ \frac{1}{t} = \lim_{t \rightarrow 0} \frac{1}{(1 + t) ^ \frac{1}{t}} = ?. 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 e
x for x near 0 is:
e ^ x = 1 + x + \frac{x ^ 2}{2!} + \frac{x ^ 3}{3!} + ...
Now, since b ^ t = e ^ {t \ln b}, 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 b
t for t near 0.
b ^ t = e ^ {t \ln b} = 1 + t \ln b + \frac{t ^ 2 \ln ^ 2 b}{2!} + \frac{t ^ 3 \ln ^ 3 t}{3!} + ....
So we have: 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 + .... 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 a
t + 1, we have:
a ^ {t + 1} = a + (a \ln a ) t + a \frac{\ln ^ 2 a}{2!} t ^ 2 + a \frac{\ln ^ 3 a}{3!}t ^ 2 + ....
Now, we'll not find the limit of:
\lim_{t \rightarrow 0} \left(1 + \alpha t + \mathcal{O}(t ^ 2) \right) ^ \frac{1}{t}, where \alpha is any constant, and \mathcal{O} (t) is the big-oh notation,
it satisfies: \lim_{t \rightarrow 0} \frac{\mathcal{O} (t ^ 2 )}{t} = 0
\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}}
= \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}
So, the term \mathcal{O} (t ^ 2) 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?
----------------------------
\lim_{t \rightarrow 0} \left( \frac{b ^ {t + 1} - a ^ {t + 1}}{b - a} \right) ^ \frac{1}{t}
= \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}}
Drop the \mathcal{O} (t ^ 2) part, we have:
= \lim_{t \rightarrow 0} \left(1 + \frac{ (b \ln b) t - (a \ln a ) t}{b - a} \right) ^ {\frac{1}{t}}
Can you go from here? :)----------------------------
Whoops, messing up with LaTeX.
Hopefully, you can get all of it, right? :)
