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