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## Sagot :

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]