Question

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

simplify the expression inside the given limit. select the correct choice below and, if
\\( \boldsymbol{\text{a.}} \\) \\( \lim_{x \to 0} \frac{1 + 5x + \sin x}{4 \cos x} = \lim_{x \to 0} \square \\) (simplify your answer.)
\\( \boldsymbol{\text{b.}} \\) the expression inside the limit cannot be simplified.

Explanation:

Step1: Recall the limit properties

We know that for a limit of a quotient, if the limit of the denominator is non - zero, we can use the property \(\lim_{x ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x ightarrow a}f(x)}{\lim_{x ightarrow a}g(x)}\) when \(\lim_{x ightarrow a}g(x) eq0\). Also, we know the basic limits \(\lim_{x ightarrow0}\sin x = 0\), \(\lim_{x ightarrow0}\cos x=\cos(0) = 1\), \(\lim_{x ightarrow0}x = 0\) and \(\lim_{x ightarrow0}c=c\) (where \(c\) is a constant).

First, let's find the limit of the denominator \(4\cos x\) as \(x ightarrow0\). Using the limit of a constant multiple and the limit of \(\cos x\): \(\lim_{x ightarrow0}4\cos x=4\lim_{x ightarrow0}\cos x\). Since \(\lim_{x ightarrow0}\cos x = 1\), then \(\lim_{x ightarrow0}4\cos x=4\times1 = 4 eq0\).

Now, for the numerator \(1 + 5x+\sin x\), we use the sum rule of limits: \(\lim_{x ightarrow0}(1 + 5x+\sin x)=\lim_{x ightarrow0}1+\lim_{x ightarrow0}5x+\lim_{x ightarrow0}\sin x\).

We know that \(\lim_{x ightarrow0}1 = 1\), \(\lim_{x ightarrow0}5x=5\lim_{x ightarrow0}x = 5\times0 = 0\), and \(\lim_{x ightarrow0}\sin x=0\). So \(\lim_{x ightarrow0}(1 + 5x+\sin x)=1 + 0+0 = 1\).

But we can also directly substitute \(x = 0\) into the function \(\frac{1 + 5x+\sin x}{4\cos x}\) because the denominator is non - zero at \(x = 0\).

Substitute \(x = 0\) into \(\frac{1+5x+\sin x}{4\cos x}\):

When \(x = 0\), \(\sin(0)=0\), \(\cos(0) = 1\), \(5x=0\). So the function becomes \(\frac{1+0 + 0}{4\times1}=\frac{1}{4}\).

But the question is about simplifying the expression inside the limit. Since the denominator \(4\cos x\) and numerator \(1 + 5x+\sin x\) don't have a common factor to cancel, but we can evaluate the limit by direct substitution (because the denominator is non - zero at \(x = 0\)). Wait, maybe the problem is to rewrite the expression? Wait, no, the function \(\frac{1 + 5x+\sin x}{4\cos x}\) can be written as \(\frac{1}{4\cos x}+\frac{5x}{4\cos x}+\frac{\sin x}{4\cos x}\), but maybe a better way: since we can directly substitute \(x = 0\) into the function (because the limit of the denominator is non - zero), the limit of the function is the value of the function at \(x = 0\) (by the theorem of limits of continuous functions, and the function \(\frac{1 + 5x+\sin x}{4\cos x}\) is continuous at \(x = 0\) because the denominator is non - zero at \(x = 0\) and the numerator is a sum of continuous functions).

Wait, the problem says "Simplify the expression inside the given limit". Let's substitute \(x = 0\) into the expression \(\frac{1 + 5x+\sin x}{4\cos x}\) to see the simplified form? Wait, no, maybe we can split the fraction:

\(\frac{1+5x+\sin x}{4\cos x}=\frac{1}{4\cos x}+\frac{5x}{4\cos x}+\frac{\sin x}{4\cos x}\), but when \(x ightarrow0\), \(\cos x ightarrow1\), so we can rewrite the expression as \(\frac{1}{4}+\frac{5x}{4}+\frac{\sin x}{4}\) (when we take the limit as \(x ightarrow0\), since \(\cos x ightarrow1\), we can approximate \(\frac{1}{4\cos x}\approx\frac{1}{4}\), \(\frac{5x}{4\cos x}\approx\frac{5x}{4}\), \(\frac{\sin x}{4\cos x}\approx\frac{\sin x}{4}\) as \(x ightarrow0\)). But actually, since the function is continuous at \(x = 0\), the limit is just the value of the function at \(x = 0\), which is \(\frac{1+0 + 0}{4\times1}=\frac{1}{4}\). But the question is about simplifying the expression inside the limit. Wait, maybe the expression can be written as \(\frac{1}{4\cos x}+\frac{5x}{4\cos x}+\frac{\sin x}{4\cos x}\), but a simpler way is to note that when we take the limit as \(x ightarrow0\), we can substitute \(x = 0\) into the expression because the denominator is non - zero at \(x = 0\). So the expression \(\frac{1 + 5x+\sin x}{4\cos x}\) at \(x = 0\) is \(\frac{1}{4}\), but the limit as \(x ightarrow0\) of the expression is the same as the value of the expression at \(x = 0\) (because it's continuous there). Wait, maybe the problem is to simplify the expression \(\frac{1 + 5x+\sin x}{4\cos x}\) by substituting the known limits of \(\sin x\) and \(\cos x\) as \(x ightarrow0\). Since \(\sin x ightarrow0\) and \(\cos x ightarrow1\) as \(x ightarrow0\), we can rewrite the expression as \(\frac{1+5x + 0}{4\times1}=\frac{1 + 5x}{4}\) (but wait, that's not correct because \(\sin x\) is part of the numerator, but as \(x ightarrow0\), \(\sin x\) approaches 0, so the numerator approaches \(1+5x\) and the denominator approaches 4. But actually, the correct simplification when taking the limit is to use the fact that we can directly substitute \(x = 0\) into the function because the denominator is non - zero at \(x = 0\). So the expression inside the limit, when simplified (by substituting the values of \(\sin x\) and \(\cos x\) as \(x ightarrow0\)) is \(\frac{1+5x+\sin x}{4\cos x}\) can be evaluated at \(x = 0\) to get \(\frac{1}{4}\), but the limit of the expression as \(x ightarrow0\) is \(\frac{1}{4}\). Wait, maybe the problem is to find the limit by simplifying the expression. Let's do it step by step.

Step1: Check the limit of the denominator

We calculate \(\lim_{x ightarrow0}4\cos x\). Using the formula \(\lim_{x ightarrow a}kf(x)=k\lim_{x ightarrow a}f(x)\) (where \(k = 4\) and \(f(x)=\cos x\)) and \(\lim_{x ightarrow0}\cos x=\cos(0) = 1\), we get \(\lim_{x ightarrow0}4\cos x=4\times1 = 4 eq0\).

Step2: Use the quotient rule for limits

Since \(\lim_{x ightarrow0}4\cos x eq0\), we can use the rule \(\lim_{x ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x ightarrow a}f(x)}{\lim_{x ightarrow a}g(x)}\), where \(f(x)=1 + 5x+\sin x\) and \(g(x)=4\cos x\).

Step3: Calculate the limit of the numerator

We use the sum rule of limits: \(\lim_{x ightarrow0}(1 + 5x+\sin x)=\lim_{x ightarrow0}1+\lim_{x ightarrow0}5x+\lim_{x ightarrow0}\sin x\).

- \(\lim_{x ightarrow0}1 = 1\) (limit of a constant)
- \(\lim_{x ightarrow0}5x=5\lim_{x ightarrow0}x = 5\times0 = 0\) (constant multiple rule and \(\lim_{x ightarrow0}x = 0\))
- \(\lim_{x ightarrow0}\sin x=0\) (basic trigonometric limit)

So \(\lim_{x ightarrow0}(1 + 5x+\sin x)=1+0 + 0=1\).

Step4: Calculate the limit of the quotient

Using the quotient rule \(\lim_{x ightarrow0}\frac{1 + 5x+\sin x}{4\cos x}=\frac{\lim_{x ightarrow0}(1 + 5x+\sin x)}{\lim_{x ightarrow0}4\cos x}=\frac{1}{4}\).

But the question is about simplifying the expression inside the limit. Since the function \(\frac{1 + 5x+\sin x}{4\cos x}\) is continuous at \(x = 0\) (because the denominator is non - zero at \(x = 0\) and the numerator is a sum of continuous functions), the limit of the function as \(x ightarrow0\) is equal to the value of the function at \(x = 0\). So we can substitute \(x = 0\) into the function to simplify the expression for the limit.

Substituting \(x = 0\) into \(\frac{1+5x+\sin x}{4\cos x}\), we get \(\frac{1+5(0)+\sin(0)}{4\cos(0)}=\frac{1 + 0+0}{4\times1}=\frac{1}{4}\). So the simplified expression inside the limit (when we evaluate it at \(x = 0\), since the limit is the same as the function value at \(x = 0\) here) is \(\frac{1}{4}\). Wait, but the problem shows an option A where we need to fill in the box. So the limit \(\lim_{x ightarrow0}\frac{1 + 5x+\sin x}{4\cos x}=\lim_{x ightarrow0}\frac{1}{4}\) (no, that's not right). Wait, no, the function \(\frac{1 + 5x+\sin x}{4\cos x}\) can be written as \(\frac{1}{4\cos x}+\frac{5x}{4\cos x}+\frac{\sin x}{4\cos x}\), but as \(x ightarrow0\), \(\cos x ightarrow1\), so \(\frac{1}{4\cos x} ightarrow\frac{1}{4}\), \(\frac{5x}{4\cos x} ightarrow0\), \(\frac{\sin x}{4\cos x} ightarrow0\). But the correct way is that since the denominator's limit is non - zero, we can directly substitute \(x = 0\) into the original function. So the expression inside the limit, when simplified (by using the fact that we can substitute \(x = 0\) because the denominator is non - zero at \(x = 0\)) gives us \(\frac{1}{4}\). So the limit \(\lim_{x ightarrow0}\frac{1 + 5x+\sin x}{4\cos x}=\lim_{x ightarrow0}\frac{1}{4}\)? No, that's incorrect. Wait, no, the original function is \(\frac{1 + 5x+\sin x}{4\cos x}\), and when \(x ightarrow0\), we can substitute \(x = 0\) into it, so the limit is \(\frac{1+0 + 0}{4\times1}=\frac{1}{4}\). So the simplified expression inside the limit (the function itself evaluated at \(x = 0\)) is \(\frac{1}{4}\), so the limit of that constant (since \(\frac{1}{4}\) is a constant) as \(x ightarrow0\) is \(\frac{1}{4}\).

Answer:

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