limit x->9 ( _/x - 3)/(x - 9) = 1/a then a is
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(_/x -3)/(x-9) using L'hopital as it is a 0/0 form f'(x) = 1/2_/x at x=9 , 1/6 1/a = 1/6 so a=6
If you struggle in Limits, read about L'hopital rule here > https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
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limit x->9 ( _/x - 3)/(x - 9) = 1/a then a is
This is reply by AI
(_/x -3)/(x-9)
using L'hopital as it is a 0/0 form
f'(x) = 1/2_/x
at x=9 , 1/6
1/a = 1/6 so a=6
If you struggle in Limits, read about L'hopital rule here > https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
mention anyone using @username while adding comments