Find $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$ . $f ( x ) = 5 \sqrt { x }$
A) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 5 } { 2 \sqrt { x } }$
B) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 7 } { 2 \sqrt { x } }$
C) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 3 } { 2 \sqrt { x } }$
D) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 5 } { 2 \sqrt { x } }$
E) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 7 } { 2 \sqrt { x } }$

Question

Find $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$ .  $f ( x ) = 5 \sqrt { x }$ 
A) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 5 } { 2 \sqrt { x } }$
B) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 7 } { 2 \sqrt { x } }$
C) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 3 } { 2 \sqrt { x } }$
D) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = \frac { 5 } { 2 \sqrt { x } }$
E) $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } = - \frac { 7 } { 2 \sqrt { x } }$

Quizplus · Accepted Answer

This limit represents the derivative of $f(x) = 5\sqrt{x}$ with respect to $x$. The derivative of $5\sqrt{x}$ using the power rule ($n\cdot x^{n-1}$) for $x^{1/2}$ is $\frac{5}{2}x^{-1/2}$, which simplifies to $\frac{5}{2\sqrt{x}}$.

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