ksaimouli one year ago limit

1. ksaimouli

$(1/x- 1/5)/(x-5)$

2. ksaimouli

as x-> 5

3. amistre64

what are the degrees of the top and bottom?

4. ksaimouli

1

5. amistre64

-1 and 1 do you recall your rules for degrees?

6. ksaimouli

What do you mean by degree? You mean power of x?

7. amistre64

yes

8. amistre64

might only work for proper polynomilas tho simplify by multiplying top and bottom by 5x

9. ksaimouli

5-x/(5x^2-25x)

10. amistre64

(5-x)/(5x(x-5)) -(x-5)/(5x(x-5)) -/(5x)

11. amistre64

-1/(5x)

12. ksaimouli

-1/25, got you

13. ksaimouli

or we could use L'hospitals rule right?

14. anonymous

Or just the definition of the derivative.

15. amistre64

the fractiony top part tends to play havok with derivatives, but its worth a shot if you are allowed to use it

16. amistre64

1/x derives to -1/x^2 ... and the limit thing is usually x to 0 tho

17. ksaimouli

By looking at it, how did you come up to multiply by 5x?

18. amistre64

/x and /5 have 5x as a common denomto clear the fractions

19. ksaimouli

Got you! thanks

20. amistre64

lhop has no issues with this

21. amistre64

good luck

22. anonymous

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: $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}~~\implies~~f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$