Math / Calculus

QuestionConsider the limit $\lim _{x \rightarrow 0} \frac{\sin ^{2}(8 x)}{1-\cos (8 x)}$
To simplify this limit, we should multiply numerator and denominator by the expression $\square$ After doing this and simplifying the result we find that the value of limit is $\square$
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Studdy Solution

STEP 1

1. The limit involves trigonometric functions and requires simplification.
2. The expression $\frac{\sin^2(8x)}{1-\cos(8x)}$ can be simplified using trigonometric identities.
3. Multiplying by a conjugate can help simplify the expression.

STEP 2

1. Identify the conjugate to multiply the expression.
2. Simplify the expression using trigonometric identities.
3. Evaluate the limit.

STEP 3

To simplify the expression, multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $1 - \cos(8x)$ is $1 + \cos(8x)$ . Therefore, multiply by:
$\frac{1 + \cos(8x)}{1 + \cos(8x)}$

STEP 4

Simplify the expression after multiplying by the conjugate:
$\frac{\sin^2(8x)(1 + \cos(8x))}{(1 - \cos(8x))(1 + \cos(8x))}$
The denominator becomes:
$(1 - \cos(8x))(1 + \cos(8x)) = 1 - \cos^2(8x) = \sin^2(8x)$
So the expression simplifies to:
$\frac{\sin^2(8x)(1 + \cos(8x))}{\sin^2(8x)}$
Cancel $\sin^2(8x)$ from the numerator and denominator:
$1 + \cos(8x)$

STEP 5

Now evaluate the limit as $x \rightarrow 0$ :
$\lim_{x \rightarrow 0} (1 + \cos(8x))$
Since $\cos(8x)$ approaches $\cos(0) = 1$ as $x \rightarrow 0$ , the limit is:
$1 + 1 = 2$
The expression to multiply by is:
$\boxed{1 + \cos(8x)}$
The value of the limit is:
$\boxed{2}$

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