Find the limit of x as it approaches zero. (Sin^2 8x)/x^2

lim ((sin 8x)^2)/x^2 as x->0

since as x->0 the denominator becomes zero, use l'hopital's rule:
lim (16 sin(8x)cos(8x) / 2x)
lim (8 sin(16x))/2x
another l'hopital's rule:
lim (128 cos (16x))/2
as x->0,
(128 * cos(0))/2
= 64

as x --> 0 sin (8x)^2---> 64 x^2

64 x^2/x^2 = 64

To find the limit of a function as it approaches a particular value, we can use algebraic manipulation and properties of limits. In this case, we will calculate the limit of the function (sin^2 8x)/x^2 as x approaches zero.

First, we notice that the function has a trigonometric expression, so we can apply a trigonometric identity to simplify it. The identity we will use is sin^2 θ = (1 - cos(2θ))/2.

Using this identity, we rewrite the function as [(1 - cos(16x))/2] / x^2.

Next, we divide the numerator and denominator by x^2 to separate the fraction and get [(1 - cos(16x))/2] / [x^2 / 1].

Now, we can split the fraction into two separate limits:

Limit as x approaches 0 of (1 - cos(16x)) / 2

and

Limit as x approaches 0 of x^2 / 1.

Let's start by calculating the limit of the first expression:

Limit as x approaches 0 of (1 - cos(16x)) / 2.

Since the function (1 - cos(16x)) is continuous at x = 0, we can directly substitute 0 into the function:

(1 - cos(16(0))) / 2 = (1 - cos(0)) / 2 = (1 - 1) / 2 = 0 /2 = 0.

So, the limit of (1 - cos(16x)) / 2 as x approaches 0 is 0.

Now, let's calculate the limit of the second expression:

Limit as x approaches 0 of x^2 / 1.

Again, we can directly substitute 0 into the function:

0^2 / 1 = 0 / 1 = 0.

Therefore, the limit of x^2 / 1 as x approaches 0 is 0.

Since both the numerator and denominator in the original expression have limits of 0 as x approaches 0, we can conclude that the limit of the function (sin^2 8x)/x^2 as x approaches 0 is also 0.