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Sagot :
To solve the limit:
[tex]\[ \lim_{x \to 64} \frac{\sqrt[6]{x} - 2}{\sqrt[3]{x} - 4} \][/tex]
we will proceed step-by-step.
1. Identify the form of the limit:
Substitute [tex]\( x = 64 \)[/tex] into the expression:
[tex]\[ \sqrt[6]{64} - 2 \quad \text{and} \quad \sqrt[3]{64} - 4 \][/tex]
- [tex]\(\sqrt[6]{64} = 2\)[/tex]
- [tex]\(\sqrt[3]{64} = 4\)[/tex]
So, substituting [tex]\( x = 64 \)[/tex] gives:
[tex]\[ \frac{2 - 2}{4 - 4} = \frac{0}{0} \][/tex]
This is an indeterminate form, which means we should use algebraic manipulation or L'Hopital's Rule to find the limit.
2. Apply L'Hopital's Rule:
Since we have a [tex]\(\frac{0}{0}\)[/tex] indeterminate form, we can use L'Hopital's Rule, which states that:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} \text{ in the form of } \frac{0}{0} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
Here,
[tex]\[ f(x) = \sqrt[6]{x} - 2 \quad \text{and} \quad g(x) = \sqrt[3]{x} - 4 \][/tex]
3. Differentiate the numerator and the denominator:
Using the power rule for derivatives, where [tex]\(\frac{d}{dx} x^n = nx^{n-1}\)[/tex]:
[tex]\[ f(x) = x^{1/6} - 2 \implies f'(x) = \frac{1}{6} x^{-5/6} \][/tex]
[tex]\[ g(x) = x^{1/3} - 4 \implies g'(x) = \frac{1}{3} x^{-2/3} \][/tex]
4. Form the new limit using the derivatives:
[tex]\[ \lim_{x \to 64} \frac{f'(x)}{g'(x)} = \lim_{x \to 64} \frac{\frac{1}{6} x^{-5/6}}{\frac{1}{3} x^{-2/3}} \][/tex]
5. Simplify the new expression:
Simplify the fraction:
[tex]\[ \lim_{x \to 64} \frac{\frac{1}{6} x^{-5/6}}{\frac{1}{3} x^{-2/3}} = \lim_{x \to 64} \frac{1}{6} x^{-5/6} \times \frac{3}{x^{-2/3}} = \lim_{x \to 64} \frac{1}{6} \times 3 \times x^{-5/6 + 2/3} \][/tex]
Since [tex]\(\frac{2}{3} = \frac{4}{6}\)[/tex], the exponents can be simplified:
[tex]\[ -\frac{5}{6} + \frac{4}{6} = -\frac{1}{6} \][/tex]
So:
[tex]\[ \lim_{x \to 64} \frac{1}{2} x^{-1/6} \][/tex]
6. Evaluate the simplified limit:
[tex]\[ = \frac{1}{2} \times (64)^{-1/6} \][/tex]
- [tex]\((64)^{-1/6}\)[/tex] is the same as [tex]\(\frac{1}{\sqrt[6]{64}}\)[/tex]
- [tex]\(\sqrt[6]{64} = 2\)[/tex]
So:
[tex]\[ (64)^{-1/6} = \frac{1}{2} \][/tex]
Therefore:
[tex]\[ \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \][/tex]
Hence, the limit is:
[tex]\[ \boxed{0.25} \][/tex]
[tex]\[ \lim_{x \to 64} \frac{\sqrt[6]{x} - 2}{\sqrt[3]{x} - 4} \][/tex]
we will proceed step-by-step.
1. Identify the form of the limit:
Substitute [tex]\( x = 64 \)[/tex] into the expression:
[tex]\[ \sqrt[6]{64} - 2 \quad \text{and} \quad \sqrt[3]{64} - 4 \][/tex]
- [tex]\(\sqrt[6]{64} = 2\)[/tex]
- [tex]\(\sqrt[3]{64} = 4\)[/tex]
So, substituting [tex]\( x = 64 \)[/tex] gives:
[tex]\[ \frac{2 - 2}{4 - 4} = \frac{0}{0} \][/tex]
This is an indeterminate form, which means we should use algebraic manipulation or L'Hopital's Rule to find the limit.
2. Apply L'Hopital's Rule:
Since we have a [tex]\(\frac{0}{0}\)[/tex] indeterminate form, we can use L'Hopital's Rule, which states that:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} \text{ in the form of } \frac{0}{0} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
Here,
[tex]\[ f(x) = \sqrt[6]{x} - 2 \quad \text{and} \quad g(x) = \sqrt[3]{x} - 4 \][/tex]
3. Differentiate the numerator and the denominator:
Using the power rule for derivatives, where [tex]\(\frac{d}{dx} x^n = nx^{n-1}\)[/tex]:
[tex]\[ f(x) = x^{1/6} - 2 \implies f'(x) = \frac{1}{6} x^{-5/6} \][/tex]
[tex]\[ g(x) = x^{1/3} - 4 \implies g'(x) = \frac{1}{3} x^{-2/3} \][/tex]
4. Form the new limit using the derivatives:
[tex]\[ \lim_{x \to 64} \frac{f'(x)}{g'(x)} = \lim_{x \to 64} \frac{\frac{1}{6} x^{-5/6}}{\frac{1}{3} x^{-2/3}} \][/tex]
5. Simplify the new expression:
Simplify the fraction:
[tex]\[ \lim_{x \to 64} \frac{\frac{1}{6} x^{-5/6}}{\frac{1}{3} x^{-2/3}} = \lim_{x \to 64} \frac{1}{6} x^{-5/6} \times \frac{3}{x^{-2/3}} = \lim_{x \to 64} \frac{1}{6} \times 3 \times x^{-5/6 + 2/3} \][/tex]
Since [tex]\(\frac{2}{3} = \frac{4}{6}\)[/tex], the exponents can be simplified:
[tex]\[ -\frac{5}{6} + \frac{4}{6} = -\frac{1}{6} \][/tex]
So:
[tex]\[ \lim_{x \to 64} \frac{1}{2} x^{-1/6} \][/tex]
6. Evaluate the simplified limit:
[tex]\[ = \frac{1}{2} \times (64)^{-1/6} \][/tex]
- [tex]\((64)^{-1/6}\)[/tex] is the same as [tex]\(\frac{1}{\sqrt[6]{64}}\)[/tex]
- [tex]\(\sqrt[6]{64} = 2\)[/tex]
So:
[tex]\[ (64)^{-1/6} = \frac{1}{2} \][/tex]
Therefore:
[tex]\[ \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \][/tex]
Hence, the limit is:
[tex]\[ \boxed{0.25} \][/tex]
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