Find limit of sin x / x^2 as x approaches 0 from the right

[hank]

Find limit of sin x / x^2 as x approaches 0 from the right

I understand that sin x / x = 1.

It seems to me that I want to make sin x / x^2 look like sin x / x.

Am I on the right track?

I imagine that you multiply sin x / x^2 by x which then gives you x * sin x / x^2. However, this just gives you x * 1 which is x. If you plug in 0 for x, you get 0, which is the wrong answer.

It should be infinity, but I'm not sure how to get there.

Where am I going wrong?

Best Regards,

--Hank
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Edited by stapel -- Reason for edit: making question visible

 

[galactus]

Are you allowed to use L"Hopital?. This is an indeterminate from.

\(\displaystyle \L\\\lim_{x\to\0}\frac{sin(x)}{x^{2}}\)

Differentiate num. and den:

\(\displaystyle \L\\\lim_{x\to\0+}\frac{cos(x)}{2x}\)

\(\displaystyle \L\\\frac{1}{2}\lim_{x\to\0+}\frac{cos(x)}{x}\)

\(\displaystyle \L\\\frac{1}{2}\lim_{x\to\0+}cos(x)\lim_{x\to\0+}\frac{1}{x}\)

Watch the 1/x. It approaches infinity as x approaches 0 from the right.

Graph it and see. It approaches negative infinity as x approaches 0 from the left.

 

[pka]

That limit does not exist. |[sin(x)/x][1/x]|= |[sin(x)/x]| |1/x|
Near zero |[sin(x)/x]| is bounded by 1, but |1/x| is not bounded.

 

[galactus]

Wow, pka, I completely overlooked that. Isn't that something. I checked using my technology and they, also, gave infinity as the answer. Tricky.

Mighty perspicacious.

EDIT:

\(\displaystyle \L\\\lim_{x\to0+}\frac{sin(x)}{x^{2}}=\lim_{x\to\0+}\frac{\frac{sin(x)}{x}}{x}={+\infty}\)

 

[hank]

Wow, pka, I completely overlooked that. Isn't that something. I checked using my technology and they, also, gave infinity as the answer. Tricky.

Mighty perspicacious.

EDIT:

\(\displaystyle \L\\\lim_{x\to0+}\frac{sin(x)}{x^{2}}=\lim_{x\to\0+}\frac{\frac{sin(x)}{x}}{x}={+\infty}\)

Oh, I see now!

Thanks tons...