Question

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

Answer 1

are r and t constants?

Answer 2

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

Answer 3

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

Answer 4

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