# 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.

• I cannot fully "parse" your limit expression. Do you mean $\lim_{t\to 0}\frac{t^2}{1-\cos^2t}$? Jan 12, 2014 at 12:50
• One round of L'Hopital gives you $\frac{0}{0}$. Jan 12, 2014 at 12:50

$\lim_{t\to\ 0} \frac{t^2}{\sin^2(t)}=\lim_{t\to\ 0}\frac{2t}{2\sin(t)\cos(t)}=\lim_{t\to 0}\frac{1}{\cos^2(t)-\sin^2(t)}=1$ So, when we apply L'Hôpital correctly we get 1.

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 $1*1$

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 $0$.

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.