Mathematics OpenStudy (anonymous):

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 OpenStudy (agreene):

are r and t constants? OpenStudy (anonymous):

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

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

$\lim_{x->0} \frac{2-(32-x)^{1/5}}{x}$ 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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