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ksaimouli
 one year ago
limit
ksaimouli
 one year ago
limit

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ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0\[(1/x 1/5)/(x5)\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2what are the degrees of the top and bottom?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.21 and 1 do you recall your rules for degrees?

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0What do you mean by degree? You mean power of x?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2might only work for proper polynomilas tho simplify by multiplying top and bottom by 5x

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2(5x)/(5x(x5)) (x5)/(5x(x5)) /(5x)

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0or we could use L'hospitals rule right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or just the definition of the derivative.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2the fractiony top part tends to play havok with derivatives, but its worth a shot if you are allowed to use it

amistre64
 one year ago
Best ResponseYou've already chosen the best response.21/x derives to 1/x^2 ... and the limit thing is usually x to 0 tho

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0By looking at it, how did you come up to multiply by 5x?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2/x and /5 have 5x as a common denomto clear the fractions

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2lhop has no issues with this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For \(h\to0\), where \(h\) represents the difference between two values of the independent variable. You can make a small modification to that definition to arrive at the derivative at a point: \[f'(x)=\lim_{h\to0}\frac{f(x+h)f(x)}{h}~~\implies~~f'(c)=\lim_{x\to c}\frac{f(x)f(c)}{xc}\]
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