Find $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$ . $f ( x ) = \frac { 1 } { x - 3 }$
A) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 4 } { ( x - 3 ) ^ { - 2 } }$
B) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 6 } { ( x - 3 ) ^ { 2 } }$
C) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 1 } { ( x - 3 ) ^ { - 2 } }$
D) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 1 } { ( x - 3 ) ^ { 2 } }$
E) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 1 } { ( x - 3 ) ^ { 2 } }$

Question

Find $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$ .  $f ( x ) = \frac { 1 } { x - 3 }$  
A) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 4 } { ( x - 3 ) ^ { - 2 } }$
B) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 6 } { ( x - 3 ) ^ { 2 } }$
C) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 1 } { ( x - 3 ) ^ { - 2 } }$
D) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 1 } { ( x - 3 ) ^ { 2 } }$
E) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 1 } { ( x - 3 ) ^ { 2 } }$

Quizplus · Accepted Answer

This limit represents the derivative of $f(x) = \frac{1}{x-3}$ with respect to $x$. The derivative of $f(x)$ is $f'(x) = -\frac{1}{(x-3)^2}$, which matches option E.

Find lim⁡h→0f(x+h)−f(x)h\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }limh→0​hf(x+h)−f(x)​

Find $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$