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I'm trying to review for my final and need help solving this problem that was given in a past homework assignment. The answer I keep getting is $-4/3$, but the correct answer is the limit DNE. I know I can approach this problem by testing different paths, but the way my professor taught it was to use the math $y = mx$. Can someone please guide me through this step-by-step?

$$\lim_{(x, y) \to (0, 0)} \frac{\sin(2x) - 2x + y}{x^3 + y}$$

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  • $\begingroup$ It would be helpful if you included the limit. The approach $y = mx$ will prove that a limit does not exists, but it does not prove when it does exist. $\endgroup$ – Doug M Dec 14 '17 at 1:01
  • $\begingroup$ You should typeset the question. In the future that link may disappear. $\endgroup$ – zhw. Dec 14 '17 at 1:37
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. It is better to type your question rather than posting an image since images cannot be searched. $\endgroup$ – N. F. Taussig Dec 14 '17 at 15:37
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$$\lim_{(x,y)\to(0,0)}\dfrac{\sin(2x)-2x+y}{x^3+y}$$

Along the line $x=0$ (y axis), we have $\dfrac{\sin(2x)-2x+y}{x^3+y}\to\dfrac{\sin(0)-0+y}{0^3+y}=\dfrac{y}{y}=1$

Along the line $y=x^3$, we have $\lim_{x\to 0}\dfrac{\sin(2x)-2x+x^3}{x^3+x^3}=\lim_{x\to 0}\dfrac{\sin(2x)-2x+x^3}{2x^3}$

Now, $$\lim_{x\to 0}\dfrac{\sin(2x)-2x+x^3}{2x^3}=\lim_{x\to 0}\dfrac{2\cos(2x)-2+3x^2}{6x^2}(LH)=\lim_{x\to 0}\dfrac{-4\sin(2x)+6x}{12x}(LH)=\lim_{x\to 0}\dfrac{-8\cos(2x)+6}{12}=\dfrac{-8+6}{12}=\dfrac{-2}{12}=\dfrac{-1}{6}$$

$(LH)$ means I applied L'Hopital rule. Clearly two different limits, therefore non exist

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  • $\begingroup$ How did you know to test the line y = x^3? $\endgroup$ – Ellie Queens Dec 14 '17 at 1:43
  • $\begingroup$ My problem with these sort of questions is knowing which paths to choose, because sometimes the paths x = 0, y = 0, and y = x all have the same result, but the path y = -x would have a different answer, making the limit DNE. But how would I know which paths to choose? $\endgroup$ – Ellie Queens Dec 14 '17 at 1:44
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    $\begingroup$ Well as you said, you always try to basic paths first (x/y axis, etc). Then if those don't work, you have to think of another path. Something I like to do is look at the "highest" power of $x$ or $y$, and basically set it accordingly. Here we have $x^3$, that's why I make $y=x^3$, because then the denominator will nicely become $2x^3$. Something you must keep in mind however is where you are going. You cannot choose something like $y=x+1$ since the point $(0,0)$ does not lie there, but in $y=-x, y=x^3$, you are fine. So yes, $y=-x$ would also work, it would give another answer, and thus DNE. $\endgroup$ – K Split X Dec 14 '17 at 1:55
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    $\begingroup$ The only way you will know is through practise of different kinds of problem $\endgroup$ – K Split X Dec 14 '17 at 1:56
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You can do the Taylor expansion on $\sin 2x$ giving $\frac {-\frac {8x^3}{6} + y}{x^3 + y}$

If we set y=0 and let x approach zero, the limit would appear to be $-\frac 43$ but if we set x = 0 and let y approach 0 the limit would appear to be $1$

$\lim_\limits{x\to 0} f(x,0) \ne \lim_\limits{y\to 0} f(0,y)$

If the limit exists, it must be the same along all paths toward 0.

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