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What is wrong with the following equation?x2+x−20/x−4=x+5a. (x−4)(x−5)≠x2+x−20b. The left-hand side is not defined for x = 0, but the right-hand side is.c. The left-hand side is not defined for x = 4, but the right-hand side is.d. None of these - the equation is correctIn view of part (a), explain why the equation lim x rightarrow 3 x2+x-12/x-3= lim x rightarrow 3 (x+4) is correct. 1. Since x2+x-12/x-3 and x + 4 are both continuous, the equation follows. 2. Since the equation holds for all x 3, it follows that both sides of the equation approach the same limit as x Rightarrow 3. This equation follows from the fact that the equation in part (a) is correct. 4. None of these 5. the equation is not correct.

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Answer:

The answer to this question can be defined as follows:

In part (i), the answer is "option d".

In part (ii), the answer is "option 2".

Step-by-step explanation:

Given:

Part (a)

[tex]\Rightarrow \bold{\frac{x^2+x-20}{x-4}=x+5}\\\\[/tex]

Solve  the above equation:

[tex]\Rightarrow x^2+x-20=(x+5)(x-4)\\\\\Rightarrow x^2+x-20=(x^2-4x+5x-20)\\\\\Rightarrow x^2+x-20=x^2-4x+5x-20\\\\\Rightarrow \boxed{x^2+x-20=x^2+x-20}\\[/tex]

Given:

Part (b)

[tex]\Rightarrow \bold{ \lim_{x \to \3} \frac{x^2+x-12}{x-3}= \lim_{x \to 3} (x+4)}\\\\[/tex]

Solve the above equation:

factor of [tex]\Rightarrow x^2+x-20 =(x-3)(x+4)[/tex]

[tex]\Rightarrow \lim_{x \to \3} \frac{(x-3)(x+4)}{x-3}= \lim_{x \to 3} (x+4)\\\\\Rightarrow \lim_{x \to \3} (x+4)= \lim_{x \to 3} (x+4)\\\\[/tex]

apply limit value:

[tex]\Rightarrow (3+4)= (3+4)\\\\\Rightarrow 7= 7[/tex]

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