Question

let f(x) = \frac{5x^{2}-29x - 6}{x^{2}+19x - 150}. find the indicated quantities, if they exist.
(a) \lim_{x\to6} f(x)
(b) \lim_{x\to0} f(x)
(c) \lim_{x\to1} f(x)
(a) select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. \lim_{x\to6} \frac{5x^{2}-29x - 6}{x^{2}+19x - 150}= (type an integer or a simplified fraction.)
b. the limit does not exist.

Explanation:

Step1: Factor the numerator and denominator

Factor $5x^{2}-29x - 6=(5x + 1)(x - 6)$ and $x^{2}+19x - 150=(x + 25)(x - 6)$. So, $f(x)=\frac{(5x + 1)(x - 6)}{(x + 25)(x - 6)}$.

Step2: Simplify the function for $x

eq6$
Cancel out the common factor $(x - 6)$ (since we are finding the limit as $x
ightarrow6$, not evaluating at $x = 6$), we get $f(x)=\frac{5x + 1}{x + 25}$ for $x
eq6$.

Step3: Find $\lim_{x

ightarrow6}f(x)$
Substitute $x = 6$ into $\frac{5x+1}{x + 25}$, we have $\frac{5\times6+1}{6 + 25}=\frac{30 + 1}{31}=\frac{31}{31}=1$.

Step4: Find $\lim_{x

ightarrow0}f(x)$
Substitute $x = 0$ into the original - simplified function $\frac{5x + 1}{x + 25}$, we get $\frac{5\times0+1}{0 + 25}=\frac{1}{25}$.

Step5: Find $\lim_{x

ightarrow1}f(x)$
Substitute $x = 1$ into the original - simplified function $\frac{5x + 1}{x + 25}$, we get $\frac{5\times1+1}{1 + 25}=\frac{6}{26}=\frac{3}{13}$.

Answer:

(A) A. $\lim_{x
ightarrow6}\frac{5x^{2}-29x - 6}{x^{2}+19x - 150}=1$
(B) $\lim_{x
ightarrow0}f(x)=\frac{1}{25}$
(C) $\lim_{x
ightarrow1}f(x)=\frac{3}{13}$