Answer:

Explanation:
![\tt \lim_(x \to 64) \frac{\sqrt[6]{x}-2 }{\sqrt[3]{x}-4 }](https://img.qammunity.org/2024/formulas/mathematics/high-school/8rp64wwhvv1o35erjzy7kmmaq1onxpdzgz.png)
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}](https://img.qammunity.org/2024/formulas/mathematics/high-school/4t798j3lhonpqd8lm2t2gf0j37r4m3z04m.png)
Using L' Hospital's rule, we need to take the derivative of the numerator and the derivative of the denominator separately.
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)}}}](https://img.qammunity.org/2024/formulas/mathematics/high-school/xmfeg06tlqhf5b3f10mfnlwa4ti1h0pghb.png)
Now we can substitute x=64 into the limit expression, and we get:
