# Homework Help: Simple limit question

1. Dec 12, 2004

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

2. Dec 12, 2004

### 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 ($\frac{0}{0}$).

Last edited: Dec 12, 2004
3. Dec 12, 2004

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

4. Dec 12, 2004

### Nylex

He means L'Hopital's rule. Mathworld can't seem to spell hopital :/, but read their page on L'Hopital's rule.

5. Dec 12, 2004

### Pyrrhus

Sorry, my native language is spanish , 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

6. Dec 12, 2004

### 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 (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.)

Last edited: Dec 12, 2004
7. Dec 12, 2004