Find the limit, if it exists.
-$$\lim _ { h \rightarrow 0 } \frac { 19 x + h } { x ^ { 3 } ( x - h ) }$$
A) Does not exist
B) $\frac { 19 } { x ^ { 4 } }$
C) $\frac { 19 } { \mathrm { x } ^ { 3 } }$
D) $19 \mathrm { x }$

Question

Quizplus · Accepted Answer

To find the limit, we can substitute x for h + x, as long as x $
eq$ 0:
$$\lim_{h ightarrow 0} \frac{19x + h}{x^3(x-h)} = \lim_{h ightarrow 0} \frac{19(x+h)}{(x+h)^3((x+h)-h)} = \lim_{h ightarrow 0}\frac{19(x+h)}{x^3+3x^2h+3xh^2+h^3-x^3} $$
$$= \lim_{h ightarrow 0} \frac{19}{x^3+3x^2h+3xh^2+h^3} = \frac{19}{x^3}$$
Therefore, the limit is $\frac{19}{x^3}$, and the answer is C.

Find the Limit, If It Exists lim⁡h→019x+hx3(x−h)\lim _ { h \rightarrow 0 } \frac { 19 x + h } { x ^ { 3 } ( x - h ) }h→0lim​x3(x−h)19x+h​

Find the Limit, If It Exists $\lim _ { h \rightarrow 0 } \frac { 19 x + h } { x ^ { 3 } ( x - h ) }$