Suppose $f ( 1 ) = 2 , \lim _ { x \rightarrow 1 ^ { - } } f ( x ) = 2 , \text { and } \lim _ { x \rightarrow 1 ^ { + } } f ( x ) = 2 , \lim _ { h \rightarrow 0 ^ { - } } \frac { f ( 1 + h ) - f ( 1 ) } { h } = - 2$ and $\lim _ { h \rightarrow 0 ^ { + } } \frac { f ( 1 + h ) - f ( 1 ) } { h } = - 2$ Is $f$ continuous and/or differentiable at $x = 1 ?$
A) f is not continuous but differentiable at x = 1
B) f is neither continuous nor differentiable at x = 1
C) f is continuous but not differentiable at x = 1
D) f is both continuous and differentiable at x = 1

Question

Quizplus · Accepted Answer

The function $f$ is continuous at $x = 1$ because the limit as $x$ approaches 1 from both sides equals the function value at $x = 1$, which is 2. It is differentiable at $x = 1$ because the limit of the difference quotient exists and is the same from both sides, equal to $-2$.

Suppose And lim⁡h→0+f(1+h)−f(1)h=−2\lim _ { h \rightarrow 0 ^ { + } } \frac { f ( 1 + h ) - f ( 1 ) } { h } = - 2limh→0+​hf(1+h)−f(1)​=−2

Suppose And $\lim _ { h \rightarrow 0 ^ { + } } \frac { f ( 1 + h ) - f ( 1 ) } { h } = - 2$