Can You Solve This Limit Definition Problem?

[Francolino]

I want to show the next statement:
$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} = \frac {1}{2} $$

What I've tried so far:
$$ \begin{align*}
\left |\frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} \right | &\leq \frac {\left |e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right ) \right |}{\left \| (x,y) \right \|^2} \\
&\leq \frac {\left |e^{x+y^2}-1 \right |+ \left |\sin \left ( x + \frac{y^2}{2} \right ) \right |}{\left \| (x,y) \right \|^2} \\
&\leq \frac {\left |x+y^2 \right | + \left | x + \frac{y^2}{2} \right |}{\left \| (x,y) \right \|^2} \\
&\leq \frac {\left \| (x,y) \right \|^2 + \left \| (x,y) \right \|^2}{\left \| (x,y) \right \|^2} \\
&\leq \boxed {2 < \delta}
\end{align*} $$
So I can take $ \boxed {\delta= \varepsilon /2} $, right? Is it well done?

I've a few more questions:
(1) Are my inequalities well done?
(2) Do you have any suggestion for this kind of problem?
(3) When I'm making these inequalities, do I'm thinking that $ x $ and $ y $ are small or big numers? Are they less or equal than 1 or bigger?

 

[zzephod]

What I've tried so far: ...

If you wish to prove that $\lim_{x,y \to 0}f(x,y)=1/2$, how does proving that $|f(x,y)|<2$ help? maybe you should be looking at $|f(x,y)-1/2|$ instead.

 

[Francolino]

If you wish to prove that $\lim_{x,y \to 0}f(x,y)=1/2$, how does proving that $|f(x,y)|<2$ help? maybe you should be looking at $|f(x,y)-1/2|$ instead.

You're right! I was reading other limit and I confused the result (the other limit was zero). I've changed the proof: $$ \begin{align*}
\left |\frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} - \frac{1}{2} \right | &\leq \frac {\left |e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right ) \right |}{\left \| (x,y) \right \|^2} + \frac{1}{2} \\
&\leq \frac {\left |e^{x+y^2}-1 \right |+ \left |\sin \left ( x + \frac{y^2}{2} \right ) \right |}{\left \| (x,y) \right \|^2} + \frac{1}{2} \\
&\leq \frac {\left |x+y^2 \right | + \left | x + \frac{y^2}{2} \right |}{\left \| (x,y) \right \|^2} + \frac{1}{2} \\
&\leq \frac {\left \| (x,y) \right \|^2 + \left \| (x,y) \right \|^2}{\left \| (x,y) \right \|^2} + \frac{1}{2} \\
&\leq \boxed {\frac{3}{2} < \delta}
\end{align*} $$
So I can take $ \boxed{ \delta = \frac {2}{3}\varepsilon} $. Is it right?

I'm a newbie, I hope you can understand my mistakes.

(Can anyone provide proofs (in book, articles, web) like the one I did? I couldn't find more than one or two examples, which were a lot easier than the limit I dealt with.)

 

[Prove It]

To prove a two variable limit $\displaystyle \begin{align*} \lim_{(x,y) \to (a,b)} f(x,y) = L \end{align*}$ by definition you need to show this:

$\displaystyle \begin{align*} \sqrt{ (x-a)^2 + (y-b)^2} < \delta \implies \left| f(x,y) - L \right| < \epsilon \end{align*}$

 

[Francolino]

To prove a two variable limit $\displaystyle \begin{align*} \lim_{(x,y) \to (a,b)} f(x,y) = L \end{align*}$ by definition you need to show this:

$\displaystyle \begin{align*} \sqrt{ (x-a)^2 + (y-b)^2} < \delta \implies \left| f(x,y) - L \right| < \epsilon \end{align*}$

Yes, so in my case $ (a,b) = (0,0) $ then, I have to show that: $$ \sqrt{x^2+y^2} =: \left \| (x,y) \right \| < \delta \Rightarrow \left |\frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} - \frac{1}{2} \right | < \varepsilon $$
And that was exactly what I was trying to show.

 

[Francolino]

Can anybody help me out with this proof?

 

[MarkFL]

Can anybody help me out with this proof?

Hello, Francolino!

Here at MHB, we discourage this kind of post, referred to as a bumping post:

No bumping. Bumping a thread is posting a reply to that thread solely to raise its profile and return it to the top of the active threads list. This is forbidden at MHB. If you want to draw attention to an unanswered thread, then post something of value such as further progress. It is also forbidden to bump one thread by drawing attention to it in a different thread.

If you have any further progress on this problem with which you need help, then it is fine to post, but simply posting to bump the thread, which adds no value to the thread, is not encouraged. :D