\lim_(x \to \y64)(x+x^(1/2)+x^(1/3)-76)/(x^(1/2)-8) Could you solve it without L'hopital rule please?

Question

`\lim_(x \to \y64)(x+x^(1/2)+x^(1/3)-76)/(x^(1/2)-8)` Could you solve it without L'hopital rule please?

Answer 1

`\lim_(x \to 64) (\frac{x+x^{(1)/(2) } +x^{(1)/(3) }-76}{x^{(1)/(2) } } } )` = 0

Explanation:

Explanation:-

Given

`\lim_(x \to 64) (\frac{x+x^{(1)/(2) } +x^{(1)/(3) }-76}{x^{(1)/(2) } } } )`

=
`(\frac{64+(64)^{(1)/(2) } +(64)^{(1)/(3) }-76}{(64)^{(1)/(2) } } } )`

=
`(\frac{64+(8^(2) )^{(1)/(2) } +(4^(3) )^{(1)/(3) }-76}{(8^(2) )^{(1)/(2) } } } )`

=
`(64+8+4-76)/(8) = (0)/(8) = 0`

Final answer:-

`\lim_(x \to 64) (\frac{x+x^{(1)/(2) } +x^{(1)/(3) }-76}{x^{(1)/(2) } } } )` = 0