# Limit

• Sep 30th 2013, 12:30 PM
abualabed
Limit
• Sep 30th 2013, 01:00 PM
topsquark
Re: Limit
What have you tried? Can you use L'Hopital's rule?

-Dan
• Sep 30th 2013, 08:44 PM
Prove It
Re: Limit
Quote:

Originally Posted by abualabed

If you're allowed to use L'Hospital's Rule, use it.

If not, rewrite it as

\displaystyle \displaystyle \begin{align*} \frac{\tan{(x)} - \sin{(x)}}{x^3} &= \frac{\tan{(x)}}{x^3} - \frac{\sin{(x)}}{x^3} \\ &= \frac{\sin{(x)}}{x^3\cos{(x)}} - \frac{\sin{(x)}}{x^3} \\ &= \frac{\sin{(x)}}{x} \cdot \frac{1}{\cos{(x)}} \cdot \frac{1}{x^2} - \frac{\sin{(x)}}{x} \cdot \frac{1}{x^2} \end{align*}

Go from there, using the well-known limit \displaystyle \displaystyle \begin{align*} \lim_{x \to 0} \frac{\sin{(x)}}{x} = 1 \end{align*}.
• Sep 30th 2013, 09:32 PM
votan
Re: Limit
Quote:

Originally Posted by abualabed

Attachment 29352
• Sep 30th 2013, 10:57 PM
abualabed
Re: Limit
hi
• Sep 30th 2013, 11:05 PM
Prove It
Re: Limit
Quote:

Originally Posted by abualabed
hi

Where on Earth did you get \displaystyle \displaystyle \begin{align*} 2 \times \frac{1}{4} \end{align*}?
• Sep 30th 2013, 11:41 PM
abualabed
Re: Limit
palestine
• Oct 1st 2013, 12:05 AM
votan
Re: Limit
Quote:

Originally Posted by abualabed
palestine

a subtle error: limit for x --> 0 not (x/2) --> 0 not the same thing.
• Oct 1st 2013, 12:24 AM
abualabed
Re: Limit
if x-->0 then (x/2)-->0
• Oct 1st 2013, 04:31 AM
votan
Re: Limit
Quote:

Originally Posted by abualabed
if x-->0 then (x/2)-->0

Which one converges faster to 0 ? this cannot be ignored in the serch for the limit of OP question.

Attachment 29356