# Squeeze Theorem Help

• Oct 9th 2010, 10:20 AM
lilaziz1
Squeeze Theorem Help
Hey everyone. I'm having some conflict with Squeeze Theorem for two variable functions. Here are a couple of examples:

$\displaystyle \lim_{(x,y)\rightarrow(0,0)} \frac{xy\cos y}{3x^2 + y^2}$

Now one can say,

$\displaystyle -1 \leq \frac{\cos y}{3x^2 + y^2} \leq 1$

$\displaystyle -xy \leq \frac{xy\cos y}{3x^2 + y^2} \leq xy$

$\displaystyle \lim_{(x,y)\rightarrow(0,0)} (-xy) = 0$

$\displaystyle \lim_{(x,y)\rightarrow(0,0)} xy = 0$

Therefore, limit exists. Or does it?

If we let $\displaystyle f(x,y) = \frac{xy\cos y}{3x^2 + y^2}$ and approach (0,0) along the line y = mx, this happens:

$\displaystyle f(x, mx) = \frac{mx^2\cos (mx)}{x^2(3 + m^2)} = \lim_{x\rightarrow0} \frac{m\cos (mx)}{3 + m^2} = \frac{m}{3 + m^2}$

∴$\displaystyle \lim_{(x,y)\rightarrow(0,0)} \frac{xy\cos y}{3x^2 + y^2}$

does not exist because one gets a different answer for all non vertical line. Thus, limit does not exist. But then why did Squeeze Theorem say it did?

Another example:

$\displaystyle \lim_{(x,y)\rightarrow(0,0)} \frac{x^2y^3}{2x^2+y^2}$

Now one can say,

$\displaystyle \frac{x^2}{2x^2+y^2} \leq 1$

Then clearly

$\displaystyle 0 \leq \frac{x^2y^3}{2x^2+y^2} \leq |y^3|$

$\displaystyle |y^3| \rightarrow 0$ as (x,y) approaches (0,0). So by the squeeze theorem,

$\displaystyle |\frac{x^2y^3}{2x^2+y^2}| \rightarrow 0$ as (x, y) approaches (0, 0)

Why does the latter example hold true while the former does not? Thank you in advance!
• Oct 9th 2010, 11:52 AM
Opalg
Quote:

Originally Posted by lilaziz1
Now one can say,

$\displaystyle -1 \leq \frac{\cos y}{3x^2 + y^2} \leq 1$

One can say it, but it isn't true. For example, if x=1/3 and y=0 then $\displaystyle \frac{\cos y}{3x^2 + y^2}$ is equal to 3.
• Oct 9th 2010, 01:27 PM
lilaziz1
:O! I get it now! This is not true: $\displaystyle 3x^2 + y^2 \geq cosy$ all the time. It makes sense. Thank you!