2. ## Re: Limit

What have you tried? Can you use L'Hopital's rule?

-Dan

3. ## Re: Limit

Originally Posted by abualabed
If you're allowed to use L'Hospital's Rule, use it.

If not, rewrite it as

\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 \begin{align*} \lim_{x \to 0} \frac{\sin{(x)}}{x} = 1 \end{align*}.

4. ## Re: Limit

Originally Posted by abualabed

hi

6. ## Re: Limit

Originally Posted by abualabed
hi
Where on Earth did you get \displaystyle \begin{align*} 2 \times \frac{1}{4} \end{align*}?

palestine

8. ## Re: Limit

Originally Posted by abualabed
palestine
a subtle error: limit for x --> 0 not (x/2) --> 0 not the same thing.

9. ## Re: Limit

if x-->0 then (x/2)-->0

10. ## Re: Limit

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.