Chapter 11.3, Problem 24E

Solution

To find: The value of $\underset{x\to 0}{\lim }\frac{3\sin x-4\cos x}{5\sin x+\cos x}$

The value is $\underset{x\to 0}{\lim }\frac{3\sin x-4\cos x}{5\sin x+\cos x}=-4$ .

Given information:

The given function is $\underset{x\to 0}{\lim }\frac{3\sin x-4\cos x}{5\sin x+\cos x}$ .

Formula used:

Difference Rule: $\underset{x\to c}{\lim }\left(f\left(x\right)-g\left(x\right)\right)=\underset{x\to c}{\lim }f\left(x\right)-\underset{x\to c}{\lim }g\left(x\right)$

Addition Rule: $\underset{x\to c}{\lim }\left(f\left(x\right)+g\left(x\right)\right)=\underset{x\to c}{\lim }f\left(x\right)+\underset{x\to c}{\lim }g\left(x\right)$

Quotient Rule: $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to c}{\lim }f\left(x\right)}{\underset{x\to c}{\lim }g\left(x\right)}$

Root Rule: $\underset{x\to c}{\lim }\sqrt[n]{f\left(x\right)}=\sqrt[n]{\underset{x\to c}{\lim }f\left(x\right)}$ , for $n\ge 2$ a positive integer.

Change of base rule: $\frac{1}{{\log }_{a}b}={\log }_{b}a$

Power rule for log: $\log m^n=n\log m$

Log of the same number as base: ${\log }_{a}a=1$

Calculation:

Substitute $0$ for $c$ , $3\sin x-4\cos x$ for $f\left(x\right)$ and $5\sin x+\cos x$ for $g\left(x\right)$ in the quotient rule formula.

  $\underset{x\to 0}{\lim }\frac{3\sin x-4\cos x}{5\sin x+\cos x}=\frac{\underset{x\to 0}{\lim }3\sin x-4\cos x}{\underset{x\to 0}{\lim }5\sin x+\cos x}$

Substitute $0$ for $c$ , $e^x$ for $f\left(x\right)$ and $\sqrt{x}$ for $g\left(x\right)$ in the difference rule formula.

  $\begin{array}{c}\underset{x\to 0}{\lim }3\sin x-4\cos x=3\sin 0-4\cos 0\\ =0-4\\ =-4\end{array}$

Substitute $0$ for $c$ , $5\sin x$ for $f\left(x\right)$ and $\cos x$ for $g\left(x\right)$ in the addition rule formula.

  $\begin{array}{c}\underset{x\to 0}{\lim }5\sin x+\cos x=\underset{x\to 0}{\lim }5\sin x+\underset{x\to 0}{\lim }\cos x\\ =5\sin 0+\cos 0\\ =0+1\\ =1\end{array}$

Therefore,

  $\begin{array}{c}\underset{x\to 0}{\lim }\frac{3\sin x-4\cos x}{5\sin x+\cos x}=\frac{-4}{1}\\ =-4\end{array}$

Hence, $\underset{x\to 0}{\lim }\frac{3\sin x-4\cos x}{5\sin x+\cos x}=-4$