Find the limit by direct substitution. $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$
A) $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$ = $\frac { 1 } { 19 }$
B) $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$ = -19
C) $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$ = $\infty$
D) $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$ = $\frac { 19 } { 2 }$
E) $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$ = $- \frac { 2 } { 19 }$

Question

Find the limit by direct substitution.   $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$  
A) $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$ = $\frac { 1 } { 19 }$
B) $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$ = -19
C) $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$ = $\infty$
D) $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$ = $\frac { 19 } { 2 }$
E) $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$ = $- \frac { 2 } { 19 }$

Quizplus · Accepted Answer

Substituting $x = 4$ directly into the expression gives $\frac{4 - 1}{4^2 + 8(4) + 9} = \frac{3}{16 + 32 + 9} = \frac{3}{57} = \frac{1}{19}$.

Find the Limit by Direct Substitution lim⁡x→4(x−1x2+8x+9)\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)limx→4​(x2+8x+9x−1​)

Find the Limit by Direct Substitution $\lim _ { x \rightarrow 4 } \left( \frac { x - 1 } { x ^ { 2 } + 8 x + 9 } \right)$