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

  • 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

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.

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    $\begingroup$ 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

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 $\sin(a)\sim_0a$.


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