0
$\begingroup$

Evaluate the limit or prove that it doesn't exist . $$\lim_{(x,y) \to (0,0)}\frac{\sin(2x) -2x + y}{x^3 + y} $$

$\endgroup$
  • 3
    $\begingroup$ Consider approaching $(0,0)$ on the line $x=0$. Then the limit (if it exists by uniqueness) goes to $1$. Now, consider the line $y=0$. Then $\lim \frac{sin(2x)-2x}{x^3} = \lim \frac{2cos(2x) - 2}{3x^2} = \lim \frac{-4sin(2x)}{6x} = \lim \frac{-8cos(2x)}{6} = -4/3$. This is nothing more than repeated use of l'Hopital's Rule. $\endgroup$ – Chris K Mar 31 '14 at 23:14
4
$\begingroup$

Take the sequence $(0,1/n)$. The limit along this sequence is $1$.

Now look at the sequence $(1/n,0)$. The limit along this sequence is $-4/3$ (Look at Taylor series expansion of sin for instance).

So, limit doesn't exist.

$\endgroup$
  • 2
    $\begingroup$ The second limit should go to $-4/3$. It is clearly negative as $\sin(x) < x$ for $x > 0$ and $\sin(x) > x$ for $x < 0$. $\endgroup$ – Chris K Mar 31 '14 at 23:16
  • $\begingroup$ @ChrisK: Thanks- you are right :) $\endgroup$ – voldemort Apr 2 '14 at 2:29
2
$\begingroup$

Hint: If the limit it exists, it exists along the line $y=x$ and it coincides with the limit along the curve $y=x^3$. Recall that $\sin(a)\sim_0a$.

$\endgroup$

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.