Question

(a) what is wrong with the following equation?
\\(\frac{x^{2}+x - 20}{x - 4}=x + 5\\)
\\((x - 4)(x + 5)\
eq x^{2}+x - 20\\)
the left - hand side is not defined for x = 0, but the right - hand side is.
the left - hand side is not defined for x = 4, but the right - hand side is.
none of these - the equation is correct.
(b) in view of part (a), explain why the following equation is correct.
\\(\lim_{x\to4}\frac{x^{2}+x - 20}{x - 4}=\lim_{x\to4}(x + 5)\\)
since \\(\frac{x^{2}+x - 20}{x - 4}\\) and x + 5 are both continuous, the equation follows.
since the equation holds for all x\\(\
eq4\\), it follows that both sides of the equation approach the same limit as x\\(\to4\\).
this equation follows from the fact that the equation in part (a) is correct.
none of these - the equation is not correct.

Explanation:

Step1: Analyze domain of left - hand side

The function $\frac{x^{2}+x - 20}{x - 4}$ has a denominator of $x - 4$. When $x=4$, the denominator is zero, and the function is undefined. The right - hand side function $y=x + 5$ is defined for all real $x$.

Step2: Analyze limit equality

Factor $x^{2}+x - 20=(x - 4)(x + 5)$. So $\frac{x^{2}+x - 20}{x - 4}=\frac{(x - 4)(x + 5)}{x - 4}=x + 5$ for $x
eq4$. As $x$ approaches 4, the limit of $\frac{x^{2}+x - 20}{x - 4}$ and $x + 5$ are the same because they are equal for all $x$ values except $x = 4$.

Answer:

(a) The left - hand side is not defined for $x = 4$, but the right - hand side is.
(b) Since the equation holds for all $x
eq4$, it follows that both sides of the equation approach the same limit as $x
ightarrow4$.