What will happen to the point ([math]x_0[/math], 0) as [math]r[/math] approaches zero? Explain your reasoning.

Now that we've made a prediction, let's see if we can prove what will happen to the point. Come up with a plan. [br][br]How can we prove what will happen to the point ([math]x_0[/math], 0) as the radius decreases to zero?

To determine what happens to the point ([math]x_0[/math], 0), we need to find the equation of line [math]PQ[/math]. In order to do this, we must identify the coordinates of points [math]P[/math] and [math]Q[/math].[br][br]What are the coordinates of point [math]P[/math]?

In order to find the coordinates of point [math]Q[/math] we must identify the equation of each circle.[br][br]Which equation describes circle [math]C_1[/math]?

Which equation describes circle [math]C_2[/math]?

What is the [i]x[/i]-coordinate of the intersection point?[br][i][br]Use the hints below if you get stuck.[/i]

Now that we have the [i]x[/i]-coordinate of the intersection point, we can find the [i]y[/i]-coordinate. Substitute the [i]x[/i]-value back into either equation and solve for [i]y[/i].[br][br]What is the [i]y[/i]-coordinate of the intersection point?[br][br][i]Use the hints below if you get stuck.[/i]

Since we have the coordinates of points [i]P[/i] and [i]Q,[/i] we can identify the slope of the line. The slope between any two points can be found using the formula [math]m=\frac{y_2-y_1}{x_2-x_1}.[/math] [br][br]     [math]P\left(0,r\right)[/math]     [math]Q\left(\frac{1}{2}r^2,\frac{1}{2}r\sqrt{4-r^2}\right)[/math][br][br]Which expression represents the slope between the two points?[br][br][i]Use the hints below if you get stuck.[/i]

Now that we have identified the slope of line [i]PQ, [/i]we can write an equation that describes it. Since we are interested in identifying the value of [math]x_0[/math], substitute [i]y[/i] = 0 into the equation and solve it for [math]x_0[/math][i].[/i][br][br]Which equation describes [math]x_0[/math]?[br][i][br]Use the hints below if you get stuck.[/i]

Now that we have an expression for [math]x_0[/math] in terms of [math]r[/math], we can evaluate its limit as [math]r[/math] approaches zero:[br][br][math]\lim_{r\to0^+}\frac{-r^2}{\sqrt{4-r^2}-2}[/math][br][br]Note that direct substitution in this instance fails, resulting in indeterminate form:[br][br][math]\lim_{r\to0^+}\frac{-r^2}{\sqrt{4-r^2}-2}=\frac{-0^2}{\sqrt{4-0^2}-2}=\frac{0}{\sqrt{4}-2}=\frac{0}{0}[/math][br][br]Our expression for [math]x_0[/math] contains a removable discontinuity. There are multiple ways we can address this. Without using calculus, we can evaluate this limit by rationalizing the denominator, using the conjugate method.[br][br]After rationalizing the denominator and simplifying the result, which of the following represents the same value as the original limit?[br][i][br]Use the hints below if you get stuck.[/i]

Now that we have removed the discontinuity, we can evaluate the limit and find the result we were looking for.[br][br]What is the result of evaluating the limit?[br][i][br]Use the hints below if you get stuck.[/i]

Was your prediction correct? Were you surprised by the results?

Before rationalizing the denominator in our limit expression we had the following:[br][br][math]\lim_{r\to0^+}\frac{-r^2}{\sqrt{4-r^2}-2}[/math][br][br]Another way to evaluate this limit is by using L'Hôpital's Rule. This rule states that to evaluate a limit that results in indeterminate form, differentiate the numerator and differentiate the denominator, and then evaluate the resulting limit.[br][br]Using L'Hôpital's Rule and the Chain Rule, evaluate the limit.[br][br][i]Click "Check" at the bottom of the page to verify all of your results.[/i]