Question

Could someone please explain how to do the following limit-problem? Evaluate the limit using the substitution z=(x^2)+(y^2) and observing that z approaches 0 from the right if and only if (x,y) approaches (0,0). (click to see limit)

Answer 1

\[\lim_{(x,y) \rightarrow (0,0)}\frac{ \sin(x^{2}+y ^{2}) }{x^{2}+y ^{2} }\]

Answer 2

@ganeshie8

Answer 3

@Jhannybean

Answer 4

@dan815

Answer 5

\[\lim_{x \rightarrow 0}\frac{ \sin x}{x}\]

Answer 6

=1

Answer 7

and how did you obtain that limit given the original?

Answer 8

i didnt, its just something you should remember

Answer 9

if you're going to make a "z" substitutuion

z= x^2 + y^2

then your limit will become

\[\lim_{(x,y) \rightarrow (0,0)}\frac{ \sin (z)}{z}\]

Answer 10

however, since the expression is now in terms of z, you need to convert the limit in relation to z

as x-> 0 and y - > 0

we also know x^2 + y^2 = z

therefore \[ z = \lim_{(x,y) \rightarrow (0,0)} x^2 + y^2\]
(note, this z is the limit, not an actual value that you will use to substitute into the substituted equation)

anyways solving this limit will tell us as x and y approaches zero, then z is approaching zero

rewriting the limit will give us

\[\lim_{z \rightarrow 0} \frac{ \sin (z) }{z}\]

Answer 11

now what i wrote previously is something that should be memorized,
\[\lim_{x \rightarrow 0} \frac{ \sin (x)}{x} = 1\]

Answer 12

please dont ask me to prove it because i dont remember the proof

Answer 13

Ok, but using l'hopital's rule works as well, correct?

Answer 14

Because you have 0/0

Answer 15

yes it does, but you really should just remember that lim x-> 0 for sin x /x = 1

Answer 16

the result is then cos(z)/1

Answer 17

which equals 1

Answer 18

its a basic identity

Answer 19

ok thanks

Answer 20

any questions?

Answer 21

no i'm ok