Simple limit question

[mad]

Hello, I tried doing this limit problem:

lim (sinx)/x^3
x-->0

here's what I did:

(lim sinx)/(lim x^3)

which should give something like 0,000000000...1 / 0,00000000001
since lim sinx (x->0) = 0 and lim x^3 = 0 too. When I make the graphic in maple, the answer is +infintiy, and the answer in my book too. Could someone clarify this for me plwase
thanks a lot

[Pyrrhus]

Do you know L'Hospital?

\lim_{x \rightarrow 0} \frac{\sin x}{x^3}

\lim_{x \rightarrow 0} \frac{\cos x}{3x^2} = \frac{1}{0} = \infty

Alternatively you could use the fact that

\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1

So we rewrite the first expression to

\lim_{x \rightarrow 0} \frac{\sin x}{x} \frac{1}{x^2}

Applying limit laws:

\lim_{x \rightarrow 0} \frac{\sin x}{x} \lim_{x \rightarrow 0} \frac{1}{x^2}

1 \cdot \frac{1}{0} = \infty

Note: when you try to calculate the limit, you should try to get it without making a indeterminate form such as the one you got ( \frac{0}{0} ).

[mad]

Do you know L'Hospital?

\lim_{x \rightarrow 0} \frac{\sin x}{x^3}

\lim_{x \rightarrow 0} \frac{\cos x}{3x^2} = \frac{1}{0} = \infty

Alternatively you could use the fact that

\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1

So we rewrite the first expression to

\lim_{x \rightarrow 0} \frac{\sin x}{x} \frac{1}{x^2}

Applying limit laws:

\lim_{x \rightarrow 0} \frac{\sin x}{x} \lim_{x \rightarrow 0} \frac{1}{x^2}

1 \cdot \frac{1}{0} = \infty

Note: when you try to calculate the limit, you should try to get it without making a indeterminate form such as the one you got ( \frac{0}{0} ).

Thank you very much for your help. I appreciate it :)
What do you mean by L'hospital? In french we say L'hopital ;).. if its a theorem we haven't seen it yet.
thanks again

[Nylex]

Thank you very much for your help. I appreciate it :)
What do you mean by L'hospital? In french we say L'hopital ;).. if its a theorem we haven't seen it yet.
thanks again

He means L'Hopital's rule. Mathworld can't seem to spell hopital :/, but read their page on L'Hopital's rule (http://mathworld.wolfram.com/LHospitalsRule.html).

[Pyrrhus]

Sorry, my native language is spanish :smile:, and i was taught L'Hospital like that, it appears like that on our books in spanish, and in the english books too, i've seen it too as L'Hospitel, oh well :smile:

[quasar987]

mad: The name of the theorem does not refer to an actual "hôpital" but as a dude whose last name was Guillaume François Antoine de L'Hospital (he was a Marquis in France I think).

Also, be careful while using those properties of the limit such as the one Cyclovenom used, that is, since

\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1

and

\lim_{x \rightarrow 0} \frac{1}{x^2} = +\infty

therefor

\lim_{x \rightarrow 0} \frac{\sin x}{x} \frac{1}{x^2} = +\infty

You have to watch out for the cases where the limits you "separated" have indeterminate forms (http://mathworld.wolfram.com/Indeterminate.html) (there are 7 of them). When you have a quotient, or product, or sum of limits that have this form, you cannot conclude to the value of limit. (as much as one would be tempted to conclude that \infty - \infty = 0 for exemple, we can not.)

[Muzza]

mad: The name of the theorem does not refer to an actual "hôpital" but as a dude whose last name was Guillaume François Antoine de L'Hospital (he was a Marquis in France I think).


This (http://scienceworld.wolfram.com/biography/LHospital.html) page explains everything.

[mad]

Thanks a lot guys!