Find $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$ . $f ( x ) = \sqrt { x - 8 }$
A) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 1 } { 2 \sqrt { x - 7 } }$
B) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 1 } { 2 \sqrt { x - 10 } }$
C) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 1 } { 2 \sqrt { x - 8 } }$
D) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 1 } { 2 \sqrt { x - 10 } }$
E) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 1 } { 2 \sqrt { x - 8 } }$

Question

Find $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$ .   $f ( x ) = \sqrt { x - 8 }$  
A) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 1 } { 2 \sqrt { x - 7 } }$
B) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 1 } { 2 \sqrt { x - 10 } }$
C) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 1 } { 2 \sqrt { x - 8 } }$
D) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 1 } { 2 \sqrt { x - 10 } }$
E) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 1 } { 2 \sqrt { x - 8 } }$

Quizplus · Accepted Answer

This limit represents the derivative of $f(x) = \sqrt{x - 8}$ with respect to $x$. The derivative of $\sqrt{x - 8}$ using the chain rule is $\frac{1}{2\sqrt{x - 8}}$, 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 $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$