limit, x -> 0, of ( ( sin(4t^2) )^2 ) / ( 5t^4 )

[calculusIsKillingMe]

lim (sin(4t²))²
x->0 (5t⁴)

 

[Deleted member 4993]

lim (sin(4t²))²
x->0 (5t⁴)

Please fix your problem in the post.There is no 'x' in \(\displaystyle \frac{sin^2(4t^2)}{5t^4}\)

Hint: Use \(\displaystyle \displaystyle{\lim_{t \to 0}\dfrac{sin(t)}{t} \ = \ 1}\)

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[Ishuda]

lim (sin(4t²))²
x->0 (5t⁴)

I assume you mean as t goes to zero [see red bold above]. What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting


Hint: Let u = 4t² and remember that
\(\displaystyle \underset{u\, \to\, 0}{lim} \frac{sin(u)}{u}\, =\, 1\)

 

[pka]

lim (sin(4t²))²
x->0 (5t⁴)

One might note that \(\displaystyle \dfrac{\sin^2(4t^2)}{5t^4}=\dfrac{16}{5}\left[\dfrac{\sin(4t^2)}{4t^2}\right]^2\)

 

[lookagain]

One might note that \(\displaystyle \dfrac{\sin(4t^2)}{5t^4}=\dfrac{16}{5}\left[\dfrac{\sin(4t^2)}{4t^2}\right]^2\)

No, it is

\(\displaystyle \dfrac{[\sin(4t^2)]^2}{5t^4} \ = \ \dfrac{16}{5}\left[\dfrac{\sin(4t^2)}{4t^2}\right]^2\)