$\text { Find } \lim _ { \Delta x \rightarrow 0 } \frac { f ( x + \Delta x ) - f ( x ) } { \Delta x } \text { where } f ( x ) = 4 x - 3 \text {. }$
A) 1
B) 4
C) $- 3$
D) 0
E) Limit does not exist.

Question

$\text { Find } \lim _ { \Delta x \rightarrow 0 } \frac { f ( x + \Delta x ) - f ( x ) } { \Delta x } \text { where } f ( x ) = 4 x - 3 \text {. }$ 
A) 1
B) 4
C) $- 3$
D) 0
E) Limit does not exist.

Quizplus · Accepted Answer

Substitute $f(x)=4x-3$ in the given limit:
$$\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x\to0}\frac{4(x+\Delta x)-3-(4x-3)}{\Delta x}$$
$$= \lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4$$
Thus, the limit exists and is equal to $\boxed{\textbf{(B) }4}$.

Find lim⁡Δx→0f(x+Δx)−f(x)Δx where f(x)=4x−3. \text { Find } \lim _ { \Delta x \rightarrow 0 } \frac { f ( x + \Delta x ) - f ( x ) } { \Delta x } \text { where } f ( x ) = 4 x - 3 \text {. } Find limΔx→0​Δxf(x+Δx)−f(x)​ where f(x)=4x−3.

$\text { Find } \lim _ { \Delta x \rightarrow 0 } \frac { f ( x + \Delta x ) - f ( x ) } { \Delta x } \text { where } f ( x ) = 4 x - 3 \text {. }$