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Sagot :
To evaluate the limit [tex]\( \lim_{x \to 0} \frac{1 - \cos(8x)}{7x^2} \)[/tex], we can use l'Hôpital's Rule because the limit presents an indeterminate form [tex]\(\frac{0}{0}\)[/tex] as [tex]\(x \to 0\)[/tex].
### Step 1: Confirm the indeterminate form
First, let's check the form of the limit:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos(8x)}{7x^2} \][/tex]
As [tex]\( x \to 0 \)[/tex]:
- [tex]\(\cos(8x) \to \cos(0) = 1\)[/tex]
- Therefore, [tex]\(1 - \cos(8x) \to 0\)[/tex]
- [tex]\(7x^2 \to 0\)[/tex]
So, the limit is of the form [tex]\(\frac{0}{0}\)[/tex], which allows us to apply l'Hôpital's Rule.
### Step 2: Apply l'Hôpital's Rule
l'Hôpital's Rule states that:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
provided that the original limit is of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex] and the limit of the derivatives exists.
Here, [tex]\( f(x) = 1 - \cos(8x) \)[/tex] and [tex]\( g(x) = 7x^2 \)[/tex].
#### Compute the derivatives:
- The derivative of the numerator [tex]\( f(x) = 1 - \cos(8x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} [1 - \cos(8x)] = 8 \sin(8x) \][/tex]
- The derivative of the denominator [tex]\( g(x) = 7x^2 \)[/tex]:
[tex]\[ g'(x) = \frac{d}{dx} [7x^2] = 14x \][/tex]
Using l'Hôpital's Rule, the limit now becomes:
[tex]\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{8 \sin(8x)}{14x} \][/tex]
### Step 3: Simplify the expression
Simplify the fraction:
[tex]\[ \frac{8 \sin(8x)}{14x} = \frac{4 \sin(8x)}{7x} \][/tex]
Now, examine the limit:
[tex]\[ \lim_{x \to 0} \frac{4 \sin(8x)}{7x} \][/tex]
### Step 4: Recognize a standard limit
Recall the standard limit:
[tex]\[ \lim_{x \to 0} \frac{\sin(kx)}{x} = k \][/tex]
where [tex]\( k \)[/tex] is a constant. In our case, [tex]\( k = 8 \)[/tex].
Therefore:
[tex]\[ \lim_{x \to 0} \frac{\sin(8x)}{x} = 8 \][/tex]
Thus:
[tex]\[ \lim_{x \to 0} \frac{4 \sin(8x)}{7x} = \frac{4}{7} \lim_{x \to 0} \frac{\sin(8x)}{x} = \frac{4}{7} \cdot 8 = \frac{32}{7} \][/tex]
### Conclusion
[tex]\[ \lim_{x \to 0} \frac{1 - \cos(8x)}{7x^2} = \frac{32}{7} \][/tex]
The evaluated limit is [tex]\( \frac{32}{7} \)[/tex].
### Step 1: Confirm the indeterminate form
First, let's check the form of the limit:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos(8x)}{7x^2} \][/tex]
As [tex]\( x \to 0 \)[/tex]:
- [tex]\(\cos(8x) \to \cos(0) = 1\)[/tex]
- Therefore, [tex]\(1 - \cos(8x) \to 0\)[/tex]
- [tex]\(7x^2 \to 0\)[/tex]
So, the limit is of the form [tex]\(\frac{0}{0}\)[/tex], which allows us to apply l'Hôpital's Rule.
### Step 2: Apply l'Hôpital's Rule
l'Hôpital's Rule states that:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
provided that the original limit is of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex] and the limit of the derivatives exists.
Here, [tex]\( f(x) = 1 - \cos(8x) \)[/tex] and [tex]\( g(x) = 7x^2 \)[/tex].
#### Compute the derivatives:
- The derivative of the numerator [tex]\( f(x) = 1 - \cos(8x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} [1 - \cos(8x)] = 8 \sin(8x) \][/tex]
- The derivative of the denominator [tex]\( g(x) = 7x^2 \)[/tex]:
[tex]\[ g'(x) = \frac{d}{dx} [7x^2] = 14x \][/tex]
Using l'Hôpital's Rule, the limit now becomes:
[tex]\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{8 \sin(8x)}{14x} \][/tex]
### Step 3: Simplify the expression
Simplify the fraction:
[tex]\[ \frac{8 \sin(8x)}{14x} = \frac{4 \sin(8x)}{7x} \][/tex]
Now, examine the limit:
[tex]\[ \lim_{x \to 0} \frac{4 \sin(8x)}{7x} \][/tex]
### Step 4: Recognize a standard limit
Recall the standard limit:
[tex]\[ \lim_{x \to 0} \frac{\sin(kx)}{x} = k \][/tex]
where [tex]\( k \)[/tex] is a constant. In our case, [tex]\( k = 8 \)[/tex].
Therefore:
[tex]\[ \lim_{x \to 0} \frac{\sin(8x)}{x} = 8 \][/tex]
Thus:
[tex]\[ \lim_{x \to 0} \frac{4 \sin(8x)}{7x} = \frac{4}{7} \lim_{x \to 0} \frac{\sin(8x)}{x} = \frac{4}{7} \cdot 8 = \frac{32}{7} \][/tex]
### Conclusion
[tex]\[ \lim_{x \to 0} \frac{1 - \cos(8x)}{7x^2} = \frac{32}{7} \][/tex]
The evaluated limit is [tex]\( \frac{32}{7} \)[/tex].
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