very simple limit question $\lim_{t\to0}\frac{t^2}{1-\cos^2t}$ I'm trying to solve some limit problems and i found this one $$ \lim_{t \to 0}\frac{t^2}{1-\cos^2t}  $$ 
what i did was, i changed $1-\cos^2t$ by $\sin^2t$ and i solve it like 
$$( \frac{t}{\sin t} \cdot \frac{t}{\sin t} )$$ and it gives me $1$.
But when i did it with L’Hospital’s Rule i get $0$.
i go with L’Hospital’s Rule, but what kind of error do i did (for 1st method) or why i get two different results?
thank you. 
 A: $\lim_{t\to\ 0} \frac{t^2}{\sin^2(t)}=\lim_{t\to\ 0}\frac{2t}{2\sin(t)\cos(t)}=\lim_{t\to 0}\frac{1}{\cos^2(t)-\sin^2(t)}=1$
So, when we apply L'Hôpital correctly we get 1.
A: if you consider that limit of product is equal to product of limit,then you should know that limit of $t/sin(t)$ using L'Hopital rule is  just $1$,therefore your limit yes it is equal to $1$
because you will have  $1*1$
A: dude , IMHO the answer is "1"
the first method you tried has been rightly executed.. your L' Hospital's theorem application has gone wrong. The numerator on differentiation gives $2t$ and the denominator leads to $(-2) \cdot (\cos t)\cdot (-\sin t )$ {which is $\sin {2t}$} or  the new limit expression is   $\frac{2t}{\sin 2t}$. Applying again L'Hospital's rule leads to the expression :  $\frac{2}{2 \cos 2t}$  or $\frac{1}{\cos 2t}$ . Now the expression is no more indeterminate at the given limit. It gives a confirming answer $1$ as $t$ tends to $0$.
A: See,
If you put the limiting value of the variable t=0 in to the function the see the function is become (0 / 0) form......
So, here you can use the l'hospital rule:  that is you can take the derivative on numerator and denominator and after applying the firt order derivative you will have something like this:  (2t) / ((2Cos t) * (Sin t)) = (2t) / Sin (2t).
 now see we have the condition as  :  llim(t--->0) (2t) / Sin (2t).
See even if you the value t=0 in this fraction the form is (0/0). So, you have to apply the derivative on numerator and denominator also for the second time!!
So, if you apply the derivative you will  get:  lim(t--->0) (2) / (2Cos (2t))  ------(a)
Now we get rid of from the 0/0 form and now we can easily put the vallue t=0 in the function (a). and we have the limiting value as 1.
Hope the discussion will help you.best of luck. 
