Find the limit $\lim _ { ( x , y ) \rightarrow ( 0,0 ) } \frac { \sin \left( 9 x ^ { 2 } + 9 y ^ { 2 } \right) } { 2 x ^ { 2 } + 2 y ^ { 2 } }$. Hint: If $x = r \cos \theta$ and $y = r \sin \theta$, then $( x , y ) \rightarrow ( 0,0 )$ if and only if $r \rightarrow 0 ^ { + }$.
A) $\frac { 9 } { 2 }$
B) 0
C) $\frac { 2 } { 9 }$
D) 1

Question

Quizplus · Accepted Answer

Using the hint and substituting $x = r \cos \theta$ and $y = r \sin \theta$, the expression becomes $\frac{\sin(9r^2)}{2r^2}$. As $r \rightarrow 0$, using the limit $\lim_{z \rightarrow 0} \frac{\sin(z)}{z} = 1$, the limit becomes $\frac{9}{2}$.

Find the Limit lim⁡(x,y)→(0,0)sin⁡(9x2+9y2)2x2+2y2\lim _ { ( x , y ) \rightarrow ( 0,0 ) } \frac { \sin \left( 9 x ^ { 2 } + 9 y ^ { 2 } \right) } { 2 x ^ { 2 } + 2 y ^ { 2 } }lim(x,y)→(0,0)​2x2+2y2sin(9x2+9y2)​

Find the Limit $\lim _ { ( x , y ) \rightarrow ( 0,0 ) } \frac { \sin \left( 9 x ^ { 2 } + 9 y ^ { 2 } \right) } { 2 x ^ { 2 } + 2 y ^ { 2 } }$