Question

let $f(x)=\frac{x^{2}-9x + 18}{x^{2}+3x - 18}$. calculate $lim_{x
ightarrow3}f(x)$ by first finding a continuous function which is equal to $f$ everywhere except $x = 3$. $lim_{x
ightarrow3}f(x)=square$

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-9x + 18=(x - 3)(x - 6)$ and the denominator $x^{2}+3x - 18=(x - 3)(x+6)$. So $f(x)=\frac{(x - 3)(x - 6)}{(x - 3)(x + 6)}$ for $x
eq3$.

Step2: Simplify the function

Cancel out the common factor $(x - 3)$ (since $x
eq3$ when taking the limit as $x
ightarrow3$). We get a new function $g(x)=\frac{x - 6}{x + 6}$, which is continuous everywhere except $x=-6$ and is equal to $f(x)$ everywhere except $x = 3$.

Step3: Calculate the limit

Substitute $x = 3$ into $g(x)$. $\lim_{x
ightarrow3}g(x)=\frac{3-6}{3 + 6}=\frac{-3}{9}=-\frac{1}{3}$.

Answer:

$-\frac{1}{3}$