0
$\begingroup$

Fix $x=0$.

$\lim _{(x,y)\to(0,0)} \tan 0 \sin {\frac{1}{\left| y \right|}}=0$.

If this limit were to exist, then it must be $0$.

Knowing that the limit does not exist, I should try other values until I reach a contradiction.

Is there a method to help me reach the contradiction faster?

$\endgroup$
1
  • $\begingroup$ $\sin$ is a bounded function, and $\tan(x) \to 0$, so squeeze theorem says.... $\endgroup$
    – user296602
    Oct 4, 2017 at 4:21

2 Answers 2

Reset to default
3
$\begingroup$

Note that $$\sin \frac{1}{|x| + |y|} \in [-1, 1]$$

for all $(x, y) \ne (0, 0)$. Now $\tan(x) \to 0$ as $x \to 0$, so squeeze theorem finishes the problem.

$\endgroup$
2
  • $\begingroup$ Sorry, I'm confused, are you trying to say that the limit is 0? Wolframalpha states the limit does not exist wolframalpha.com/input/… $\endgroup$
    – Omar Chehab
    Oct 4, 2017 at 4:38
  • 1
    $\begingroup$ Yes, the limit is zero. Wolfram is pretty sketchy with multivariable limits under the best of circumstances; compare this to this. $\endgroup$
    – user296602
    Oct 4, 2017 at 4:49
0
$\begingroup$

i would have multiplied and divided by $|x||y|$ and let $u= |x||y|$ you'll soon discover that you have: $$u\tan(x) \frac{sin u}{u}$$ the latter goes to 1 while the former goes to zero as tan 0 = 0 and 0*0*0=0

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .