2 분 소요

# Question

## Question 1

(a) What is wrong with the following equation?

$\frac{x^{2}+x-6}{x-2} = x + 3$

(b) In view of part (a), explain why the equation

$\lim_{x\to{2}}\frac{x^{2} + x - 6}{x - 2} = \lim_{x\to{2}}(x + 3)$

is correct.

## Question 2

Is there a number $a$ such that

$\lim_{x\to{-2}}{\frac{3x^{2}+ax+a+3}{x^{2}+x-2}}$

exists? If so, find the value of $a$ and the value of the limit.

## Question 3

The figure shows a fixed circle $C_{1}$ with equation $(x-1)^{2} + y^{2} = 1$ and a shrinking circle $C_{2}$ with radius $r$ and center the origin. $P$ is the point $(0, r)$, $Q$ is the upper point of intersection of the two circles, and $R$ is the point of intersection of the line $PQ$ and the x-axis. What happens to $R$ as $C_{2}$ shrinks, that is, as $r\to{0^{+}}$?

(a)

$\frac{x^{2}+x-6}{x-2} \neq x + 3$ $(\because \frac{x^{2}+x-6}{x-2} \text{ is not defined when x = 2})$

(b)

$\lim_{x\to{2}}\frac{x^{2} + x - 6}{x - 2} = \lim_{x\to{2}}\frac{(x-2)(x+3)}{x - 2}$ $= \lim_{x\to{2}}(x + 3)$ $(\because x\to{2} \text{ is not equal to }x=2)$

$\text{Since the denominator approaches 0 as } x\to{-2}$ $\text{The limit will exist only if the nominator approaches 0 as } x\to{-2}$

$\lim_{x\to{-2}}{(3x^{2}+ax+a+3)} = 0$ $3(-2)^2+a(-2)+a+3 = 0$ $\therefore a = 15$

$\text{With } a = 15, \text{the limit becomes}$ $\lim_{x\to{-2}}{\frac{3x^{2}+15x+18}{x^{2}+x-2}} = \lim_{x\to{-2}}{\frac{3(x+2)(x+3)}{(x-1)(x+2)}}$ $= \lim_{x\to{-2}}{\frac{3(x+3)}{(x-1)}}$ $(\because x\to{-2} \text{ is not equal to }x = -2)$ $= \frac{3((-2)+3)}{((-2)-1)}$ $= \frac{3}{(-3)} = -1$

$C_{1}: (x-1)^{2}+y^{2} = 1$ $C_{2}: x^{2} + y^{2} = r$ $\text{We already know the coordinates of the point P is } (0, r) \text{ from graph.}$ $\text{Thus, we must first find the coordinates of the point } Q \text{ to get the equation of the line } PR.$ $Q \text{ is the upper point of intersection of the two circles.}$

$\therefore x^{2} + 1 - (x-1)^{2} = r^{2}$ $x = \frac{r^{2}}{2}$ $x = \frac{r^{2}}{2} \text{ is the } x\text{-coordinate of the point Q.}$ $\text{Substituting back into any equation of the two circles to find the } y\text{-coordinate}.$ $(\frac{r^{2}}{2})^{2} + y^{2} = r^{2}$ $y^{2} = r^{2} - (\frac{r^{2}}{2})^{2}$ $y = \sqrt{r^{2}-(\frac{r^{2}}{2})^{2}}$ $(\because y \geq 0)$ $\therefore y = \sqrt{r^{2}-(\frac{r^{2}}{2})^{2}} \text{ is the } y\text{-coordinate of the point Q.}$ $\text{Now, The equation of the line joining } P \text{ and } Q \text{ is thus}$ $y = -\frac{r-\sqrt{r^{2}-(\frac{r^{2}}{2})^{2}}}{\frac{r^{2}}{2}}x + r$ $\text{We set } y = 0 \text{ in order to find the x-intercept, and get}$ $-\frac{r-\sqrt{r^{2}-(\frac{r^{2}}{2})^{2}}}{\frac{r^{2}}{2}}x + r = 0$ $x = \frac{-r^{3}}{-2r+2\sqrt{r^{2}-(\frac{r^{2}}{2})^{2}}}$

$\text{Let's take the limit as } r\to{0^{+}}.$ $\lim_{r\to{0^{+}}}\frac{-r^{3}}{-2r+2\sqrt{r^{2}-(\frac{r^{2}}{2})^{2}}}$

$\lim_{r\to{0^{+}}}\frac{-r^{3}}{-2r+2\sqrt{r^{2}-(\frac{r^{2}}{2})^{2}}} = \lim_{r\to{0^{+}}}\frac{-r^{3}}{-2r(1-\sqrt{1-\frac{r^{2}}{4}})}$ $= \lim_{r\to{0^{+}}}\frac{r^{2}}{2(1-\sqrt{1-\frac{r^{2}}{4}})} = \lim_{r\to{0^{+}}}\frac{r^{2}(1+\sqrt{1-\frac{r^{2}}{4}})}{2(1-\sqrt{1-\frac{r^{2}}{4}})(1+\sqrt{1-\frac{r^{2}}{4}})}$ $= \lim_{r\to{0^{+}}}\frac{r^{2}(1+\sqrt{1-\frac{r^{2}}{4}})}{2\{1-(1-\frac{r^{2}}{4})\}} = \lim_{r\to{0^{+}}}\frac{r^{2}(1+\sqrt{1-\frac{r^{2}}{4}})}{\frac{r^{2}}{2}}$ $= \lim_{r\to{0^{+}}}2(1+\sqrt{1-\frac{r^{2}}{4}}) = 4$

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