Question

can anyone please help me solve this 2 problems:
http://www.imagebam.com/image/769c24412008603

Answer 1

Have you tried evaluating limit as x and y approaches 0?

Answer 2

yes , but in the first problem i couldn't evaluate it because of the upper part of the fraction containig sin3(x^2+y^2)
lim (x,y)--->(0,0) and let y=mx

Answer 3

recall the single variable limit
\[\lim\limits_{t\to 0}~\frac{\sin t}{t}=1\]

Answer 4

along \(y=mx\) path we get :

\[\begin{align}\lim\limits_{(x,y)\to(0,0)}~\frac{\sin3(x^2+y^2)}{x^2+y^2}
&=\lim\limits_{(x,~mx)\to(0,0)}~\frac{\sin3(x^2+m^2x^2)}{x^2+m^2x^2}\\~\\
&=3*\lim\limits_{(x,~mx)\to(0,0)}~\frac{\sin3(x^2+m^2x^2)}{3(x^2+m^2x^2)}\\~\\
&=3*1
\end{align}\]

Answer 5

.

Answer 6

IrishBoy123:??

Answer 7

To me, if you let x = r cos theta
y = r sin theta
then x^2 +y^2 = r^2
then when x, y approach (0,0) , r approaches 0
\(lim_{r\rightarrow 0} \dfrac{sin 3 r^2 }{r^2}=\lim_{r\rightarrow 0}\dfrac{3sin(3r^2)}{3r^2}=3\)