Exploring the Limit of x^2sin(e/x)

[lovely_aly]

Homework Statement

Find the lim x-->0 x^2sin(e/x).

Homework Equations

The Attempt at a Solution

Using the limit laws I can say that
lim x-->0 x^2sin(e/x) = lim x-->0 x^2 * lim x-->0 sin(e/x)
lim x-->0 x^2=0; but lim x-->0 sin(e/x) however, is undefined...right?

So, the limit doesn't exist?

 

[quasar987]

The theorem is... IF lim f and lim g exist, then lim(f+g) exists and equals limf+limg. Here, lim sin(e/x) doesn not exist, so you can't use this theorem.

But try to use the facts that |sin(y)|<=1 and the fact that lim |f(x)|=0 <==> lim f(x)=0.

 

[quantumdude]

Using the limit laws I can say that
lim x-->0 x^2sin(e/x) = lim x-->0 x^2 * lim x-->0 sin(e/x)
lim x-->0 x^2=0; but lim x-->0 sin(e/x) however, is undefined...right?

No, you can't say that. You have to read the limit law carefully. It only allows you to split the limits up like that when both limits do exist.

Instead try the following change of variables. Let $u=\frac{1}{x}$. Then $x\rightarrow 0 \Rightarrow u\rightarrow\infty$.

 

[lovely_aly]

Instead try the following change of variables. Let $u=\frac{1}{x}$. Then $x\rightarrow 0 \Rightarrow u\rightarrow\infty$.

Could you clarify this a bit more for me, please? Should I be subsituting $u=\frac{1}{x}$ for (e/x)? And essentially by taking the sin(infinity) the limit would be -1? Am I following this correctly?

Forgive me if I'm a little slow with this. I'm a newbie to calculus, this is the first term I've ever taken and it's an online accelerated course. So, while I enjoy math and usually get it pretty quickly, my extreme overload of information has got my brain sort of going in circles.

And, thank you! I really appreciate the help. =)

 

[nicksauce]

He means that you should consider
$$\lim_{u\rightarrow\infty}\frac{1}{u^2}sin(ue)$$

You can then apply the squeeze theorem to this limit."And essentially by taking the sin(infinity) the limit would be -1?"
Not sure where you get this. sin(infinity) is not well-defined in any sense.

 

[HallsofIvy]

You might also use the fact that
$\lim_{x\rightarrow 0}\frac{sin(x)}{x}= 1$

 

[JinM]

Halls: can't he use squeeze theorem directly without the substitution, or am I missing something?

 

[HallsofIvy]

Yes, you could. I was just suggesting another possibility.