limit

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[(1/x- 1/5)/(x-5)\]
as x-> 5
what are the degrees of the top and bottom?

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Other answers:

1
-1 and 1 do you recall your rules for degrees?
What do you mean by degree? You mean power of x?
yes
might only work for proper polynomilas tho simplify by multiplying top and bottom by 5x
5-x/(5x^2-25x)
(5-x)/(5x(x-5)) -(x-5)/(5x(x-5)) -/(5x)
-1/(5x)
-1/25, got you
or we could use L'hospitals rule right?
Or just the definition of the derivative.
the fractiony top part tends to play havok with derivatives, but its worth a shot if you are allowed to use it
1/x derives to -1/x^2 ... and the limit thing is usually x to 0 tho
By looking at it, how did you come up to multiply by 5x?
/x and /5 have 5x as a common denomto clear the fractions
Got you! thanks
lhop has no issues with this
good luck
For \(h\to0\), 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: \[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}~~\implies~~f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\]

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