Multivariable Calc

[meks0899]

I just started my multivariable calc class last week. We were given soem problems to try. See attached picture for the problem. We were told to consider 3 cases. ? < ?, ? > ?, and ? = ?. I managed to get the first two but I don't know how to do the last case which involves ? = ?.

 

[mmm4444bot]

Hi meks,

Proofs are not my forte, but here's my reasoning. You can decide if there's a better way of expressing it.

If alpha = beta, then we can say the following:

lim[x --> infty] x_n = alpha

lim[x --> infty] y_n = alpha

Therefore, in the limit:

x_n = y_n

This means, in the limit:

max{x_n, y_n} = x_n

because this is actually max{x_n, x_n}

And, of course, we can say:

max{alpha, beta} = alpha

This means that the original limit to prove:

lim[x --> infty] max{x_n, y_n} = max{alpha, beta}

is the same (when n is infinite) to one of the given limits:

lim[x --> infty] x_n = alpha

Does that work? Perhaps, it could be stated more concisely.

The same line of reasoning can proceed using max{beta, beta} and max{y_n, y_n} to show that the limit to prove is the same as the given limit for y_n.

Cheers

~ Mark

 

[daon]

Thought this thread could use a little more rigor

Let $\displaystyle \epsilon > 0$.

First things first: If f and g are any positive real-valued functions,

We can find a $\displaystyle N_1 \in \mathbb{N}$ such that $\displaystyle |x_n-\alpha| < f(\epsilon)$ whenever $\displaystyle n>N_1$ [1]
We can find a $\displaystyle N_2 \in \mathbb{N}$ such that $\displaystyle |y_n-\beta| < g(\epsilon)$ whenever $\displaystyle n>N_2$ [2]

If $\displaystyle \alpha = \beta$ then Let $\displaystyle N = \max\{N_1, N_2\}$ with $\displaystyle f=g=\epsilon$.

If $\displaystyle n > N$ then [1] and [2] both hold. Therefore $\displaystyle |\max\{x_n,y_n\} - \alpha| = |\max\{x_n,y_n\} - \max\{\alpha, \beta\}| < \epsilon$

Now assume WLOG, $\displaystyle \alpha > \beta$. You wish to show $\displaystyle \max\{x_n,y_n\} \to \alpha$ as $\displaystyle n \to \infty$.

Suppose $\displaystyle f=g=\min\{\frac{\alpha-\beta}{2}, \epsilon\}$

Let $\displaystyle N = \max \{N_1,N_2\}$ as above.

Then from [1] & [2] with our choice above for f and g, we get: (I'll leave the details and algebra to you)

$\displaystyle y_n < \beta + \frac{\alpha-\beta}{2}= \alpha - \frac{\alpha-\beta}{2} < x_n$

Thus $\displaystyle n > N \implies x_n = \max\{x_n,y_n\}$

The result follows.