# How do you find the limit of #sqrt(1- 8x^3) /sqrt(1-4x^2)# as x approaches 1/2?

We will use the following identities, which are a difference of cubes and a difference of squares.

Thus, we have the limit

Square roots can be split up through multiplication:

graph{sqrt(1-8x^3)/sqrt(1-4x^2) [-2.827, 2.648, -0.39, 2.347]}

Note:

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To find the limit of sqrt(1-8x^3) / sqrt(1-4x^2) as x approaches 1/2, we can simplify the expression by factoring out the common terms inside the square roots.

First, let's factor out the common term inside the first square root: sqrt(1-8x^3) = sqrt(8x^3-1)

Next, let's factor out the common term inside the second square root: sqrt(1-4x^2) = sqrt(4x^2-1)

Now, we can rewrite the expression as: sqrt(8x^3-1) / sqrt(4x^2-1)

To evaluate the limit as x approaches 1/2, we substitute x = 1/2 into the expression: sqrt(8(1/2)^3-1) / sqrt(4(1/2)^2-1)

Simplifying further: sqrt(1-1) / sqrt(1-1)

Since both the numerator and denominator are equal to zero, we have an indeterminate form. To resolve this, we can use L'Hôpital's rule.

Differentiating the numerator and denominator with respect to x: d/dx (sqrt(8x^3-1)) / d/dx (sqrt(4x^2-1))

Simplifying the derivatives: (12x^2) / (4x)

Substituting x = 1/2 into the derivatives: (12(1/2)^2) / (4(1/2))

Simplifying further: (12/4) / (4/2)

Final simplification: 3/2

Therefore, the limit of sqrt(1-8x^3) / sqrt(1-4x^2) as x approaches 1/2 is 3/2.

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To find the limit of ( \frac{\sqrt{1-8x^3}}{\sqrt{1-4x^2}} ) as ( x ) approaches ( \frac{1}{2} ), you can use L'Hôpital's Rule or algebraic simplification.

Applying algebraic simplification, you can factor out ( \sqrt{1} ) from each expression, simplify, and then substitute ( x = \frac{1}{2} ).

( \lim_{{x \to \frac{1}{2}}} \frac{\sqrt{1-8x^3}}{\sqrt{1-4x^2}} = \lim_{{x \to \frac{1}{2}}} \frac{\sqrt{1(1-8x^3)}}{\sqrt{1(1-4x^2)}} = \lim_{{x \to \frac{1}{2}}} \frac{\sqrt{1(1-8x^3)}}{\sqrt{1(1-4x^2)}} = \lim_{{x \to \frac{1}{2}}} \frac{\sqrt{1-4x^2}}{\sqrt{1-4x^2}} = \frac{\sqrt{1-4(1/2)^2}}{\sqrt{1-4(1/2)^2}} = \frac{\sqrt{1-1}}{\sqrt{1-1}} = \frac{0}{0} )

As the denominator and the numerator both approach zero, you can apply L'Hôpital's Rule by finding the derivative of both the numerator and denominator with respect to ( x ) and then taking the limit.

( \lim_{{x \to \frac{1}{2}}} \frac{\sqrt{1-8x^3}}{\sqrt{1-4x^2}} = \lim_{{x \to \frac{1}{2}}} \frac{\frac{d}{dx} \sqrt{1-8x^3}}{\frac{d}{dx} \sqrt{1-4x^2}} )

After finding the derivatives, substitute ( x = \frac{1}{2} ) and evaluate the limit.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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