Analysis: finding limit of tough sequence

[K29]

Homework Statement

Use bernouillis inequality to show that
$$\stackrel{lim}{_{n \rightarrow \infty}} (\frac{1+\frac{x+y}{n}}{1+\frac{x+y}{n}+\frac{xy}{n^{2}}})^{n}=1$$
$$x,y \in R$$

Homework Equations

The Attempt at a Solution

With simple manipulation this equals:

$$\stackrel{lim}{_{n \rightarrow \infty}}(1-\frac{\frac{xy}{n^{2}}}{1+\frac{x+y}{n}+\frac{xy}{n^{2}}})^{n}$$

For sufficiently large n I can apply Bernoullis inequality and get:
$$\geq \stackrel{lim}{_{n \rightarrow \infty}}1-\frac{\frac{xy}{n}}{1+\frac{x+y}{n}+\frac{xy}{n^{2}}}$$

Which equals 1.

Now I was hoping to squeeze my original sequence between this and something else, but I don't know what that something else could be. Is this the right way of doing things? Help please

 

[Dick]

I'd look at the numerator and denominator of your original expression separately. The numerator is (1+(x+y)/n)^n. Take the log and use l'Hopital.

 

[K29]

yeah, I know that would slide if this were a calculus course or something. But we can't use log because we haven't done the analysis of the limit n(x^(1/n)-1) yet. which is natural log, and the analysis of real functions like log or exponentation of real numbers or anything like that would be way beyond the scope of any undergrad real analysis course

e.g. https://www.physicsforums.com/showthread.php?t=266960

In terms of real analysis that would be cheating until you have defined and done the analysis of real functions and done a lot of analysis to do with differentiation to build all the way up to l'hospital's rule.
We had to prove that by doing substitutions and creating decreasing functions that tend to the same limit to prove it is a finite limit as euler may have done I suppose.

I assume this is why we must use Bernoullis inequality instead of using maths that follows from real analysis.

 

[lepton5]

I think yes that limit eqn can be solved elegantly using Bernoulli ineq.

suppose $$p = 1 + \frac{x+y}{n}$$, subsitute it to the original eqn, and you can get:

$$(\frac{1}{1+\frac{xy}{p n^2}})^n$$. Apply Bernoulli ineq to denominator:

$$(1+\frac{xy}{p n^2})^n = 1 + \frac{xy}{pn}$$. input back to limit,

$$\stackrel{lim}{_{n \rightarrow \infty}} (\frac{1}{1 + \frac{xy}{pn}}) = 1$$

 

[K29]

Nice. I assume you meant $$(1+ \frac{xy}{pn^{2}})^{n} \geq 1+\frac{xy}{pn}$$ for sufficiently large n.
Which gives me the other side of the inequality to squeeze the original limit between the one I originally worked out and this one. Thanks.