Find the limit for the given function. lim t-->0 (sin(8t))^2/t^2
as t-->0, we have 8t-->0, so
lim (sin(8t))^2/t^2 =
[lim sin(8t)/t].[limsin(8t)/t] as8t-->0
=8.[limsin(8t)/8t].8[limsin(8t)/8t], 8t-->0
= 8*1*8*1= 64
you know the formula lim (sint/t)=1 as
t-->1, aren't you?
yea i knw... thnxx foh ur help
To find the limit of the function as t approaches 0, we can use the properties of limits and basic trigonometric identities.
Let's simplify the expression a bit before finding the limit. We know that (sin x)^2 = (1/2)(1 - cos(2x)). Therefore, we can rewrite the given function as:
lim t->0 [(1/2)(1 - cos(16t)) / t^2]
Now, let's split the expression into two limits, as the limit of a quotient is equivalent to the quotient of the limits:
lim t->0 (1/2) * lim t->0 (1 - cos(16t)) / lim t->0 t^2
The first limit can be evaluated separately:
lim t->0 (1 - cos(16t))
As t approaches 0, the cosine function oscillates between 1 and -1. Therefore, evaluating this limit at t = 0 gives us:
1 - cos(16 * 0) = 1 - cos(0) = 1 - 1 = 0
So, the first limit simplifies to 0.
Now, let's evaluate the second limit:
lim t->0 t^2
Since t^2 is a polynomial function, the limit of t^2 as t approaches 0 is simply 0^2 = 0.
Therefore, the second limit is 0.
Now, bringing it all together:
lim t->0 (1/2) * lim t->0 (1 - cos(16t)) / lim t->0 t^2
= (1/2) * 0 / 0 = 0
Thus, the limit of the given function as t approaches 0 is 0.