MATH 116 SUMMER 2008 QUIZ 1

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Problem: Prove the following

lim_{(x,y)→(1,0)} cos (\frac{xy(x − 1)^{2}}{(x − 1)^{2} + y^{2}} ) = 1

Solution: Since cos z is continuous at z = 0 and cos 0 = 1 then it is enough to show that

lim_{(x,y)→(1,0)} \frac{xy(x − 1)^{2}}{(x − 1)^{2} + y^{2}} = 0.

Solution1: Another observation is

lim_{(x,y)→(1,0)} \frac{xy(x − 1)^{2}}{(x − 1)^{2} + y^{2}} = \{ lim_{(x,y)→(1,0)} x\}·\{ lim_{(x,y)→(1,0)} \frac{y(x − 1)^{2}}{(x − 1)^{2} + y^{2}} \} = lim_{(x,y)→(1,0)} \frac{y(x − 1)^{2}}{(x − 1)^{2} + y^{2}} .

We use ϵ − δ definition to prove that the later limit is 0. For any ϵ > 0 we have to find δ > 0 such that

0 < \sqrt{(x − 1)^{2} + y^{2}} < δ ⟹ |\frac{y(x − 1)^{2}}{(x − 1)^{2} + y^{2}} − 0| < ϵ(∗)

We have,

|\frac{y(x − 1)^{2}}{(x − 1)^{2} + y^{2}} − 0| = |y| |\frac{(x − 1)^{2}}{(x − 1)^{2} + y^{2}} | ≤ |y| = \sqrt{y^{2}} ≤ \sqrt{(x − 1)^{2} + y^{2}} < δ = ϵ

Taking δ = ϵ provides implication (∗)

Solution2: We have,

|\frac{xy(x − 1)^{2}}{(x − 1)^{2} + y^{2}} | ≤ |xy| .

Since

− |xy| ≤ \frac{xy(x − 1)^{2}}{(x − 1)^{2} + y^{2}} ≤ |xy| and lim_{(x,y)→(1,0)} − |xy| = lim_{(x,y)→(1,0)} |xy| = 0,

then, by Sandwich Theorem,

lim_{(x,y)→(1,0)} \frac{y(x − 1)^{2}}{(x − 1)^{2} + y^{2}} = 0.