Find the limit by direct substitution. $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$
A) $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$ = $\infty$
B) $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$ = $- \frac { 16 } { 17 }$
C) $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$ = $\frac { 13 } { 17 }$
D) $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$ = $- \frac { 17 } { 13 }$
E) $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$ = $\frac { 17 } { 16 }$

Question

Find the limit by direct substitution.   $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$  
A) $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$ = $\infty$
B) $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$ = $- \frac { 16 } { 17 }$
C) $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$ = $\frac { 13 } { 17 }$
D) $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$ = $- \frac { 17 } { 13 }$
E) $\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)$ = $\frac { 17 } { 16 }$

Quizplus · Accepted Answer

Substituting $x = -4$ directly into the expression gives $\frac{4(-4)}{(-4)^2 + 1} = \frac{-16}{16 + 1} = -\frac{16}{17}$.

Find the Limit by Direct Substitution lim⁡x→−4(4xx2+1)\lim _ { x \rightarrow - 4 } \left( \frac { 4 x } { x ^ { 2 } + 1 } \right)limx→−4​(x2+14x​)

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