## dgamma3 2 years ago limit question. lim (x - > 5) 1/(x-5)^4

1. dgamma3

$\lim_{x \rightarrow 5} 1/(1+5)^4$

2. dgamma3

woops. the 1 is an x

3. kirbykirby

It would be infinity

4. kirbykirby

oh

5. dgamma3

why so?

6. kirbykirby

Do you know L'Hopital's rule

7. dgamma3

nope. is it the only way?

8. dgamma3

the answer is infinity. just have no clue how to work it out.

9. kirbykirby

Ok what you can do is:

10. kirbykirby

wait i just noticed

11. kirbykirby

u mean it's x-5 in the denominator right? otherwise it wouldn't be infinity

12. dgamma3

yes, x-5

13. kirbykirby

$\lim_{x \rightarrow 5} \frac{x}{(x-5)^4}$$=\lim_{x \rightarrow 5} \frac{x}{(x-5)^4}$$=\lim_{x \rightarrow 5} \frac{x}{(x-5)^2(x-5)^2}$$=\lim_{x \rightarrow 5} \frac{x}{(x-5)(x-5)(x-5)(x-5)}$$=\lim_{x \rightarrow 5} \frac{x/x}{(x/x-5/x)(x/x-5/x)(x/x-5/x)(x/x-5/x)}$$=\lim_{x \rightarrow 5} \frac{1}{(1-5/x)(1-5/x)(1-5/x)(1-5)}$

14. kirbykirby

oops the last factor should also be (1-5/x)

15. dgamma3

awesome, thanks for your effort! just wondering line 6. why are you diving everything by x?

16. kirbykirby

I hope you see that the bottom factors are (1-1) when substituting x=5, so you get like "1/0" -> infinity

17. kirbykirby

hm it's true that was a useless step

18. kirbykirby

I was thinking that it was going to x->infinity at first, but i realized it was x->5 after lol

19. dgamma3

ahhh, yes. i see that. thank you. is that a legal move though? because you are changing the function?

20. dgamma3

for example, (x-5) doesn't equal (1-5/x)

21. dgamma3

sorry, i guess it is. just would have never thought of that!

22. kirbykirby

Nope :) If you divide the top and bottom by x, it is perfectly valid (it's like you are dividing by "1") Example: $\frac{2}{3} = \frac{\frac{2}{x}}{\frac{3}{x}}=\frac{2}{x}*\frac{x}{3}=\frac{2}{3}$

23. kirbykirby

I divided the numerator by x as well: I did x/x = 1 in the numerator, and divided the whole denominator by x

24. kirbykirby

if you divide the numerator, you MUST also divide the denominator!

25. dgamma3

awesome, this is just amazing. never would have figured out this.

26. kirbykirby

you will learn a neat theorem called "L'hopital's rule" later which will make the computation even easier :)

27. dgamma3

k. thanks for your time. im now 'a fan' haha.

28. kirbykirby

Hehe awesome ;) Good luck with everything

29. dgamma3

cheers