limit question. lim (x - > 5) 1/(x-5)^4

- anonymous

limit question. lim (x - > 5) 1/(x-5)^4

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

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

- anonymous

woops. the 1 is an x

- kirbykirby

It would be infinity

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- kirbykirby

oh

- anonymous

why so?

- kirbykirby

Do you know L'Hopital's rule

- anonymous

nope. is it the only way?

- anonymous

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

- kirbykirby

Ok what you can do is:

- kirbykirby

wait i just noticed

- kirbykirby

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

- anonymous

yes, x-5

- kirbykirby

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

- kirbykirby

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

- anonymous

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

- kirbykirby

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

- kirbykirby

hm it's true that was a useless step

- kirbykirby

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

- anonymous

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

- anonymous

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

- anonymous

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

- kirbykirby

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

- kirbykirby

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

- kirbykirby

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

- anonymous

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

- kirbykirby

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

- anonymous

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

- kirbykirby

Hehe awesome ;) Good luck with everything

- anonymous

cheers

Looking for something else?

Not the answer you are looking for? Search for more explanations.