5
$\begingroup$

$$ \lim\limits_{(x, y)\to (0, 0)}\frac{\sin(x+y)}{x+y} $$

I did the following

$a)$ along $x$ axis, the limit is one

$b)$ along $y$ axis the limit is one

$c)$ along $y=x$ the limit is one

Since there exists more ways to approach the origin, I know I cannot conclude from the steps given above.

$d)$ along $y= -x$

$\frac{\sin(x-x)}{x-x}$ is not defined. Isn't it still possible for the function to have a limit, even though it is not defined at that point?

How do I conclude whether a limit exists or doesn't in such a case?

EDIT

So for, $$ \lim\limits_{(x, y)\to (0, 0)}\frac{\sin(xy)}{xy} $$ Can I proceed in a similar manner and perform a substitution and state the limit is 1?

$\endgroup$
  • 1
    $\begingroup$ Did you try letting $(x+y) = u$ and taking $\lim_{u\to0} f(u)$? And this limit is 1 by the way. $\endgroup$ – SalmonKiller May 10 '15 at 5:00
  • $\begingroup$ Why is it reasonable to make that substitution? $\endgroup$ – getafix May 10 '15 at 5:04
  • $\begingroup$ @getaflix see my answer. $\endgroup$ – SalmonKiller May 10 '15 at 5:05
  • $\begingroup$ The substitution is reasonable because no matter how $x+y$ changes, it changes in exactly the same way everywhere. $\endgroup$ – abiessu May 10 '15 at 5:07
1
$\begingroup$

Let $x+y=t$ so that $t\to 0 $ as $(x, y) \to (0,0)$ therefore we have that $$\lim_{(x, y) \to (0,0)}\frac{\sin (x+y)}{x+y}=\lim_{t \to 0}\frac{\sin t}{t}=1$$

$\endgroup$
0
$\begingroup$

So $\lim_{(x,y) \to (0,0)} \frac{\sin(x+y)}{x+y} = 1$, since this is a simple trig limit and the form is $\frac{\sin(0)}{0}$. You might substitute $u = x+y$, and $\lim_{(x,y)\to(0,0)} x+y$ is obviously 0. Doing the substitution, we get $\lim_{u\to0} \frac{\sin u}{u} = 1$.

$\endgroup$
  • $\begingroup$ Unfortunately, we haven't covered such substitutions in our course yet and I am not sure if I am allowed to use them. $\endgroup$ – getafix May 10 '15 at 5:06
  • $\begingroup$ But, I guess I could check if 1 is indeed the limit using the epsilon-delta definition of a limit $\endgroup$ – getafix May 10 '15 at 5:07
  • $\begingroup$ @getafix substitution is a fancy way of making it simple. You can also perform the same trick without substitution, just use $x+y$ instead. $\endgroup$ – SalmonKiller May 10 '15 at 5:07
  • 2
    $\begingroup$ The substitution technique is not a valid proof. It prove the limit exists as $x+y$ approaches 0, not $(x,y) \rightarrow (0,0).$ See an earlier post for a better proof. \math.stackexchange.com/questions/291538/… $\endgroup$ – matt biesecker May 10 '15 at 5:13
  • $\begingroup$ @mattbiesecker I don't really get that proof. He says we have to "make" the function continuous, but the for the limit to exist, we need the function exists on an open interval around 0 (in this case). $\endgroup$ – SalmonKiller May 10 '15 at 5:16
0
$\begingroup$

$$\lim\limits_{(x, y)\to(0, 0)}\frac{\sin(x+y)}{x+y}$$ The Taylor series of $\sin(x+y)$ at around $(0, 0)$ is $$ (x+y)-\frac16 (x+y)^3+O\left((x+y)^5\right) $$ Therefore $$\lim\limits_{(x, y)\to(0, 0)}\frac{\sin(x+y)}{x+y}$$ $$=\lim\limits_{(x, y)\to(0, 0)}\frac{(x+y)-\frac16 (x+y)^3+O\left((x+y)^5\right)}{x+y}$$ $$=\lim\limits_{(x, y)\to(0, 0)}\left[1-\frac16 (x+y)^2+O\left((x+y)^4\right)\right]=1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.