Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$ occur frequently in calculus. Evaluate this limit for the given value of $x$ and function $f$.
-$$( x ) = \frac { 2 } { x } , x = - 5$$
A) 10
B) $- \frac { 2 } { 25 }$
C) Does not exist
D) $- \frac { 2 } { 5 }$

Question

Quizplus · Accepted Answer

The given limit represents the derivative of $f(x) = \frac{2}{x}$ at $x = -5$. To find this derivative, we use the definition of the derivative: $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$. Substituting $f(x) = \frac{2}{x}$ into this formula gives $\lim _ { h \rightarrow 0 } \frac { \frac{2}{-5+h} - \frac{2}{-5} } { h }$. Simplifying this expression and evaluating the limit yields $- \frac { 2 } { 25 }$.

Because of Their Connection with Secant Lines, Tangents, and Instantaneous 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)​

Because of Their Connection with Secant Lines, Tangents, and Instantaneous $\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$