Find the following limit, if it exists.
$$\lim _ { x \rightarrow - 3 } \frac { 9 - 3 x - 2 x ^ { 2 } } { 3 + x }$$
A) $- 3$
B) 9
C) 7
D) $- 11$
E) The limit does not exist.

Question

Quizplus · Accepted Answer

We can use direct substitution to see that the numerator of the fraction becomes 27 when x is -3, and the denominator becomes 0. Therefore, we have an indeterminate form of $\frac{0}{0}$. To remove the indeterminate form, we can factor the numerator as follows:

$$9-3x-2x^2=(3+x)(3-2x)$$

Now, we can cancel the factor of $(3+x)$ in both the numerator and denominator, and we are left with:

$$\lim_{x\rightarrow -3}\frac{9-3x-2x^2}{3+x}=\lim_{x\rightarrow -3}\frac{(3+x)(3-2x)}{3+x}=\lim_{x\rightarrow -3}(3-2x)=-9+6=\boxed{-3}$$

Find the Following Limit, If It Exists lim⁡x→−39−3x−2x23+x\lim _ { x \rightarrow - 3 } \frac { 9 - 3 x - 2 x ^ { 2 } } { 3 + x }x→−3lim​3+x9−3x−2x2​

Find the Following Limit, If It Exists $\lim _ { x \rightarrow - 3 } \frac { 9 - 3 x - 2 x ^ { 2 } } { 3 + x }$