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3 votes

\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?

User Medriscoll
by
5.0k points

1 Answer

5 votes

Answer:


\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

User Hert
by
4.9k points
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