Trig Limits Problem

[gadav478]

Good afternoon all.

I am having a problem solving this trig limit. I was hoping to be pointed in the right direction.

Problem: lim as t -> 0 (t^2)/(1-cos^2t)

I know that 1-cos^2t simplifies to sin^2t but I get stuck there. Should I separate (t^2)/(sin^2t) and get
(lim as t -> 0 (t/sint)) times (lim as t -> 0 (t/sint)) and evaluate there?

By the way, the answer is 1.

I really appreciate any assistance with understanding how this works out.

Thanks!

 

[DrPhil]

Good afternoon all.

I am having a problem solving this trig limit. I was hoping to be pointed in the right direction.

Problem: lim as t -> 0 (t^2)/(1-cos^2t)

I know that 1-cos^2t simplifies to sin^2t but I get stuck there. Should I separate (t^2)/(sin^2t) and get
(lim as t -> 0 (t/sint)) times (lim as t -> 0 (t/sint)) and evaluate there?

By the way, the answer is 1.

I really appreciate any assistance with understanding how this works out.

Thanks!

Well done.

The limit of the square is the square of the limit:
since you know that the limit of t/sin(t) as t-->0 is 1,
then the limit of (t/sin(t))^2 is (1)^2 = 1

 

[srmichael]

Good afternoon all.

I am having a problem solving this trig limit. I was hoping to be pointed in the right direction.

Problem: lim as t -> 0 (t^2)/(1-cos^2t)

I know that 1-cos^2t simplifies to sin^2t but I get stuck there. Should I separate (t^2)/(sin^2t) and get
(lim as t -> 0 (t/sint)) times (lim as t -> 0 (t/sint)) and evaluate there?

By the way, the answer is 1.

I really appreciate any assistance with understanding how this works out.

Thanks!

Usually when you see a limit problem with sin as t-->0, they are tring to push this formula on you:

\(\displaystyle \lim_{t\rightarrow0}\frac{\sin(t)}{t}=1\)

You have simplified to \(\displaystyle \frac{t^2}{sin^2(t)}\) correctly. Do you see how this can be rewritten as:

\(\displaystyle \frac{t}{\sin(t)}\cdot\frac{t}{\sin(t)}\)

Can you now take it from here?

 

[gadav478]

So DrPhil, just so I'm understanding correctly, t/sint = 1?

I know sinx/x is 1, but is this true the other way around (i.e. x/sinx)?

Thanks for the fast replies, I appreciate it very much.

 

[DrPhil]

So DrPhil, just so I'm understanding correctly, t/sint = 1?

I know sinx/x is 1, but is this true the other way around (i.e. x/sinx)?

Thanks for the fast replies, I appreciate it very much.

I am not always completely rigorous in my approach to mathematical proofs! The same procedure that you used to prove sin(x)/x --> 1 can be used for x/sin(x). I would just say
1 / 1 = 1

 

[JeffM]

Dr. Phil has given you the common sense answer.

Being more formal, there are two general theorems applicable here. One is:

\(\displaystyle [1]\ \displaystyle \lim_{x \rightarrow a}f(x) = b\ and\ \lim_{x \rightarrow}g(x) = c \implies \lim_{x \rightarrow a}\left\{f(x) * g(x)\right\} = b * c.\)

From this you can trivially deduce

\(\displaystyle [1a]\ \displaystyle \lim_{x \rightarrow d}f(x) = e \implies \lim_{x \rightarrow d}\left\{f(x)\right\}^2 = e^2.\)

The other is:

\(\displaystyle [2]\ \displaystyle \lim_{x \rightarrow p}f(x) = q \ne 0\ and\ \lim_{x \rightarrow p}g(x) = r \implies \lim_{x \rightarrow p}\dfrac{g(x)}{f(x)} = \dfrac{r}{q}.\)

So being somewhat formal

\(\displaystyle h(t) = \dfrac{t^2}{1 -cos^2(t)} = \dfrac{t^2}{sin^2(t)} = \dfrac{1}{\dfrac{sin^2(t)}{t^2}} = \left(\dfrac{1}{\dfrac{sin(t)}{t}}\right)^2\ if\ t \ne 0 \ne sin(t).\)

\(\displaystyle \displaystyle \lim_{t \rightarrow 0}\dfrac{sin(t)}{t} = 1 \implies \lim_{x \rightarrow 0}\dfrac{1}{\dfrac{sin(t)}{t}} = \dfrac{1}{1} = 1\) by Theorem 2.

\(\displaystyle \displaystyle \lim_{x \rightarrow 0}\left(\dfrac{1}{\dfrac{sin(t)}{t}}\right)^2 = 1^2 = 1\) by Theorem 1a.

\(\displaystyle So\ \displaystyle \lim_{x \rightarrow 0}\dfrac{t^2}{1 - cos^2(t)}= 1\).