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Lim
\frac{ \sqrt[6]{x} - 2}{ \sqrt[3]{x} - 4}

x-->64

Please help!​

1 Answer

1 vote

Answer:


\boxed{\tt (1)/(4)}

Explanation:


\tt \lim_(x \to 64) \frac{\sqrt[6]{x}-2 }{\sqrt[3]{x}-4 }

let's make the root in terms of Power.


\tt \lim_(x \to 64) \frac{\sqrt[6]{x}-2 }{\sqrt[3]{x}-4 } = \lim_(x \to 64) \frac{x^{(1)/(6)}-2}{x^{(1)/(3)}-4}

Using L' Hospital's rule, we need to take the derivative of the numerator and the derivative of the denominator separately.


  • \tt (d)/(dx) \left[ x^{(1)/(6)}-2 \right] = (1)/(6)x^{(1)/(6)-1}-0 = (1)/(6)x^{(-5)/(6)}= \frac{1}{6(x^{(5)/(6)})}

  • \tt (d)/(dx) \left[ x^{(1)/(3)}-4 \right] = (1)/(3)x^{(-2)/(3)} = \frac{1}{3(x^{(2)/(3)})}

So, the limit expression becomes:


\tt \lim_(x \to 64) \frac{(d)/(dx) \left[ x^{(1)/(6)}-2 \right]}{(d)/(dx) \left[ x^{(1)/(3)}-4 \right]} = \lim_(x \to 64) \frac{\frac{1}{6(x^{(5)/(6)})}}{\frac{1}{3*x^{(2)/(3)}}}

Now we can substitute x=64 into the limit expression, and we get:


\tt \lim_(x \to 64) \frac{\frac{1}{6(x^{(5)/(6)})}}{\frac{1}{3*x^{(2)/(3)}}}=\frac{3*64^{(2)/(3)}}{6(64^{(5)/(6)})}= (3 \cdot 4^2)/(6 \cdot 2^5) = (48)/(192) = \boxed{(1)/(4)}

User Andrew Rayner
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