Does the Limit Exist? Check to Find Out!

[mathmari]

Hey!

I have found applying De L'Hoptal's Rule that $$\lim_{x \rightarrow 0} \frac{\sin 2x-2x}{x^3}=-\frac{4}{3}$$

Now I am asked whether the limit $$\lim_{(x, y) \rightarrow (0, 0)} \frac{\sin 2x-2x+y}{x^3+y}$$ or not.

How could we check that ?? (Wondering)

 

[I like Serena]

Hey!

I have found applying De L'Hoptal's Rule that $$\lim_{x \rightarrow 0} \frac{\sin 2x-2x}{x^3}=-\frac{4}{3}$$

Now I am asked whether the limit $$\lim_{(x, y) \rightarrow (0, 0)} \frac{\sin 2x-2x+y}{x^3+y}$$ or not.

How could we check that ?? (Wondering)

Hi! (Wave)

I would start with checking what the limit is on different lines, say $y=0$ respectively $x=0$. (Thinking)

 

[chisigma]

Hey!

I have found applying De L'Hoptal's Rule that $$\lim_{x \rightarrow 0} \frac{\sin 2x-2x}{x^3}=-\frac{4}{3}$$

Now I am asked whether the limit $$\lim_{(x, y) \rightarrow (0, 0)} \frac{\sin 2x-2x+y}{x^3+y}$$ or not.

How could we check that ?? (Wondering)

If $(x,y) \rightarrow (0,0)$ along the trajectory y=0 the limit is $- \frac{4}{3}$ as You have found... if $(x,y) \rightarrow (0,0)$ along the trajectory $y = 2\ x - \sin 2x$ the limit is 0... what do You can conclude?...

Kind regards

$\chi$ $\sigma$