To evaluate the limit using l'Hôpital's Rule, follow these steps:
Consider the limit:
[tex]\[
\lim _{x \rightarrow 0} \frac{6 \sin 7 x}{5 x}
\][/tex]
First, check if using l'Hôpital's Rule is appropriate. L'Hôpital's Rule can be applied to the limit if it is in the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex].
### Step 1: Identify the Indeterminate Form
Evaluate the limit's numerator and denominator at [tex]\(x = 0\)[/tex]:
- [tex]\(\sin(7 \cdot 0) = \sin(0) = 0\)[/tex]
- The numerator [tex]\(6 \cdot 0 = 0\)[/tex]
- The denominator [tex]\(5 \cdot 0 = 0\)[/tex]
Since both the numerator and denominator approach 0 as [tex]\(x \to 0\)[/tex], the limit is initially in the indeterminate form [tex]\(\frac{0}{0}\)[/tex].
### Step 2: Apply l'Hôpital's Rule
L'Hôpital's Rule states that:
[tex]\[
\lim _{x \rightarrow c} \frac{f(x)}{g(x)} = \lim _{x \rightarrow c} \frac{f'(x)}{g'(x)}
\][/tex]
provided that the limit on the right-hand side exists.
Rewrite the given limit using the derivatives of the numerator and the denominator:
- Let [tex]\( f(x) = 6 \sin(7x) \)[/tex]
- Let [tex]\( g(x) = 5x \)[/tex]
Compute the derivatives:
- [tex]\( f'(x) = 6 \cdot 7 \cos(7x) = 42 \cos(7x) \)[/tex]
- [tex]\( g'(x) = 5 \)[/tex]
Thus, the limit becomes:
[tex]\[
\lim _{x \rightarrow 0} \frac{6 \sin 7 x}{5 x} = \lim _{x \rightarrow 0} \frac{42 \cos(7x)}{5}
\][/tex]
### Step 3: Evaluate the New Limit
Now, evaluate the limit of the new expression as [tex]\(x \to 0\)[/tex]:
- As [tex]\( x \to 0 \)[/tex], [tex]\(\cos(7x) \to \cos(0) = 1\)[/tex]
Therefore:
[tex]\[
\lim _{x \rightarrow 0} \frac{42 \cos(7x)}{5} = \frac{42 \cdot 1}{5} = \frac{42}{5} = 8.4
\][/tex]
So, the evaluated limit is:
[tex]\[
\lim _{x \rightarrow 0} \frac{6 \sin 7 x}{5 x} = 8.4
\][/tex]
