evaluate the limit. (if an answer does not exist, enter dne.)\n lim_{x\rightarrow3}\frac{4x^{2}-9x…

evaluate the limit. (if an answer does not exist, enter dne.)\n lim_{x\rightarrow3}\frac{4x^{2}-9x - 9}{2x^{2}-18}

evaluate the limit. (if an answer does not exist, enter dne.)\n lim_{x\rightarrow3}\frac{4x^{2}-9x - 9}{2x^{2}-18}

Answer

Explanation:

Step1: Factor the numerator and denominator

Factor $4x^{2}-9x - 9=(4x + 3)(x - 3)$ and $2x^{2}-18 = 2(x^{2}-9)=2(x + 3)(x - 3)$. So the limit becomes $\lim_{x\rightarrow3}\frac{(4x + 3)(x - 3)}{2(x + 3)(x - 3)}$.

Step2: Cancel out the common factor

Cancel out the common factor $(x - 3)$ (since $x\neq3$ when taking the limit), we get $\lim_{x\rightarrow3}\frac{4x+3}{2(x + 3)}$.

Step3: Substitute $x = 3$

Substitute $x = 3$ into $\frac{4x+3}{2(x + 3)}$, we have $\frac{4\times3+3}{2\times(3 + 3)}=\frac{12 + 3}{2\times6}=\frac{15}{12}=\frac{5}{4}$.

Answer:

$\frac{5}{4}$