Question

find the limit.
$lim_{x \to 0} \frac{1 + 5x + \sin x}{4 \cos x}$

Explanation:

Step1: Substitute \( x = 0 \) into the function

We know that for a limit as \( x \to a \) of a function \( f(x) \), if \( f(x) \) is continuous at \( x = a \), we can directly substitute \( x = a \) into \( f(x) \). The function \( \frac{1 + 5x+\sin x}{4\cos x} \) is continuous at \( x = 0 \) because the denominator \( 4\cos x \) at \( x = 0 \) is \( 4\cos(0)=4\times1 = 4
eq0 \), and the numerator is a combination of continuous functions (polynomial and sine function) at \( x = 0 \).

Substitute \( x = 0 \) into the numerator \( 1 + 5x+\sin x \):
\( 1+5\times0+\sin(0)=1 + 0+0 = 1 \)

Substitute \( x = 0 \) into the denominator \( 4\cos x \):
\( 4\cos(0)=4\times1 = 4 \)

Step2: Calculate the limit

The limit of a quotient is the quotient of the limits (when the limit of the denominator is non - zero). So \( \lim_{x\to0}\frac{1 + 5x+\sin x}{4\cos x}=\frac{\lim_{x\to0}(1 + 5x+\sin x)}{\lim_{x\to0}(4\cos x)} \)

We already found that \( \lim_{x\to0}(1 + 5x+\sin x)=1 \) and \( \lim_{x\to0}(4\cos x) = 4 \)

So \( \frac{1}{4} \)

Answer:

\(\frac{1}{4}\)