Continuous?

Calculus Level 3

The function

f ( x , y ) = { sin ⁡ ( y ) − y x 2 + y 2 ( x , y ) ≠ ( 0 , 0 ) 0 ( x , y ) = ( 0 , 0 ) f(x,y)=\begin{cases} \dfrac{\sin(y) - y}{x^2 +y^2 }& \quad (x,y) \neq (0,0) \\ \quad \quad 0 & \quad (x,y)=(0,0) \end{cases}

is continuous at ( 0 , 0 ) . (0,0).

Try this

True False

ابراهيم فقرا by ابراهيم فقرا , 23, Israel

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2 solutions

Otto Bretscher
Dec 21, 2018

lim ⁡ y → 0 sin ⁡ ( y ) − y y 2 = 0 \lim_{y \to 0} \frac{\sin (y) -y}{y^2}=0 by Bernoulli's rule so lim ⁡ ( x , y ) → ( 0 , 0 ) sin ⁡ ( y ) − y x 2 + y 2 = 0 \lim_{(x,y) \to (0,0)} \frac{\sin (y) -y}{x^2+y^2}=0 since ∣ sin ⁡ ( y ) − y x 2 + y 2 ∣ ≤ ∣ sin ⁡ ( y ) − y y 2 ∣ \left|{\frac{\sin (y) -y}{x^2+y^2}}\right|\leq \left|{\frac{\sin (y) -y}{y^2}}\right| if y ≠ 0 y\neq 0 . The claim is T r u e \boxed{True}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Taylor series of sin ⁡ ( y ) \sin(y) around y = 0 y=0 is y − y 3 6 + y 5 120 + O ( y 7 ) y-\frac{y^3}{6}+\frac{y^5}{120}+O(y^7) .

So we will get sin ⁡ ( y ) − y x 2 + y 2 = − y 3 6 + y 5 120 + O ( y 7 ) x 2 + y 2 \frac{\sin(y)-y}{x^2+y^2}=\frac{\frac{-y^3}{6}+\frac{y^5}{120}+O(y^7) }{x^2+y^2} ... it is easy to continue from here

2 Helpful 0 Interesting 2 Brilliant 1 Confused

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