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

Question

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

Quizplus · Accepted Answer

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

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