Question

homework2: problem 15 (1 point) evaluate $lim_{x \to 3} \frac{sin(x - 3)}{x^{2}+3x - 18}$. limit = you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor page generated september 28, 2025, 9:54:29 pm cdt webwork © 1996 - 2024 | theme: math4 - ttu | ww_version: 2.19 | pg_version: 2.19 the webwork project

Explanation:

Step1: Factor the denominator

Factor $x^{2}+3x - 18=(x - 3)(x+6)$. So the limit becomes $\lim_{x
ightarrow3}\frac{\sin(x - 3)}{(x - 3)(x + 6)}$.

Step2: Use the limit - property $\lim_{u

ightarrow0}\frac{\sin u}{u}=1$
Let $u=x - 3$. As $x
ightarrow3$, then $u
ightarrow0$. The limit is $\lim_{u
ightarrow0}\frac{\sin u}{u(u + 9)}$. We can rewrite it as $\lim_{u
ightarrow0}\frac{\sin u}{u}\cdot\lim_{u
ightarrow0}\frac{1}{u + 9}$.

Step3: Evaluate the limits

We know that $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$ and $\lim_{u
ightarrow0}\frac{1}{u + 9}=\frac{1}{0 + 9}=\frac{1}{9}$. Then the product of the two limits is $1\times\frac{1}{9}=\frac{1}{9}$.

Answer:

$\frac{1}{9}$