Evaluating Limit: x to Infinity

[haoku]

I am having difficulty in evaluating the limit
x tends to positive infinity (log( (1+[sqrt(2)/2])^x + (1-[sqrt(2)/2])^x ))/x

I have tried using binomial series expansion but turn out to be something messy.

Any ideas on it?
:)

 

[Gib Z]

Well.

You have
$$\lim_{x\to \infty} \frac{ \log_e \left( \left( 1 + \frac{1}{\sqrt{2}} \right)^x + \left( 1 - \frac{1}{\sqrt{2}} \right)^x \right)}{x}$$

As x tends to infinity, the second term inside the natural log goes to zero, as so we can exclude that from our limit so that it easily evaluates to $$\log_e \left( 1 + \frac{1}{\sqrt{2}} \right)$$.

This gives us the value of the limit, but really is just a heuristic argument. For some rigor, further investigation is required. Guided by our previous evaluation of the limit, we can see that it would be helpful to isolate the important term as such:

$$\log_e \left( \left( 1 + \frac{1}{\sqrt{2}} \right)^x + \left( 1 - \frac{1}{\sqrt{2}} \right)^x \right) = \log_e \left( 1 + \frac{1}{\sqrt{2}} \right)^x + \log_e \left( 1 + (3-2\sqrt{2})^x \right)$$ using log ab = log a + log b.

One can easily show the second term goes to zero, using a Taylor expansion if required.

 

[nikkor180]

Greetings:

The previous poster was indeed correct in that the numerator's second term goes to zero. That said, the limit becomes,

limit(x-->inf) [(ln(1 + 1/sqrt(2))^x) / x]. From the property log(u^n) = n*log(u), we have,

limit [x*ln(1 + 1/sqrt(2)) / x] = limit [ln(1 + 1/sqrt(2))] = ln(1 + 1/sqrt(2)) [limit of a constant is the constant].

Regards,

Rich B.
rmath4u2@aol.com