Question

In Problems 53-56, use the properties of limits and the facts that

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\sin x}{x}=1 \\ \underset{x\to 0}{\lim }\frac{\cos x-1}{x}=1 \\ \underset{x\to 0}{\lim }\sin x=0 \\ \underset{x\to 0}{\lim }\cos x=1 \end{gathered}$$

$\underset{x\to 0}{\lim }\frac{3\sin x+\cos x-1}{4x}$

Short answer

The limit equals$\frac{3}{4}$

Step-by-step solution

Given information

Given function$\underset{x\to 0}{\lim }\frac{3\sin x+\cos x-1}{4x}$

Splitting the expression and calculatinglimx→03sinx+cosx-14x=limx→03sinx4x+cosx-14xlimx→03sinx+cosx-14x=limx→03sinx4x+cosx-14xlimx→03sinx+cosx-14x=limx→034·sinxx+14·cosx-1xlimx→03sinx+cosx-14x=limx→034·sinxx+14·cosx-1xlimx→03sinx+cosx-14x=limx→034·limx→0sinxx+limx→014·limx→0cosx-1xlimx→03sinx+cosx-14x=34·1+14·0limx→03sinx+cosx-14x=34

The limit is$\frac{3}{4}$