limx->0 (2-5rt(32-x))/x...... is the answer -15/2? if not could you tell me what i did wrong? limx->0 (2-5rt(32-x))/x...... is the answer -15/2? if not could you tell me what i did wrong? @Mathematics

are r and t constants?

oo that is root 5 so (32-x)^(1/5)

oh, that changes a lot of what i was just typing, lol... lemme look at it again :P

\[\lim_{x->0} \frac{2-(32-x)^{1/5}}{x}\] use L'Hospital's rule: \[= \lim_{x->0} \frac{1}{5 (32-x)^{4/5}}\] Factor out constants: = 1/5 (lim_(x->0) 1/(32-x)^(4/5)) The limit of a quotient is the quotient of the limits: = 1/(5 (lim_(x->0) (32-x)^(4/5))) Using the power law, write lim_(x->0) (32-x)^(4/5) as (lim_(x->0) (32-x))^(4/5): \[= \frac{1}{5 [\lim_{x->0} (32-x)^{4/5}]}\] The limit of 32-x as x approaches 0 is 32: = 1/80

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