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

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\[\lim_{x \rightarrow 5} 1/(1+5)^4\]
woops. the 1 is an x
It would be infinity

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Other answers:

why so?
Do you know L'Hopital's rule
nope. is it the only way?
the answer is infinity. just have no clue how to work it out.
Ok what you can do is:
wait i just noticed
u mean it's x-5 in the denominator right? otherwise it wouldn't be infinity
yes, x-5
\[\lim_{x \rightarrow 5} \frac{x}{(x-5)^4}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x-5)^4}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x-5)^2(x-5)^2}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x-5)(x-5)(x-5)(x-5)}\]\[=\lim_{x \rightarrow 5} \frac{x/x}{(x/x-5/x)(x/x-5/x)(x/x-5/x)(x/x-5/x)}\]\[=\lim_{x \rightarrow 5} \frac{1}{(1-5/x)(1-5/x)(1-5/x)(1-5)}\]
oops the last factor should also be (1-5/x)
awesome, thanks for your effort! just wondering line 6. why are you diving everything by x?
I hope you see that the bottom factors are (1-1) when substituting x=5, so you get like "1/0" -> infinity
hm it's true that was a useless step
I was thinking that it was going to x->infinity at first, but i realized it was x->5 after lol
ahhh, yes. i see that. thank you. is that a legal move though? because you are changing the function?
for example, (x-5) doesn't equal (1-5/x)
sorry, i guess it is. just would have never thought of that!
Nope :) If you divide the top and bottom by x, it is perfectly valid (it's like you are dividing by "1") Example: \[\frac{2}{3} = \frac{\frac{2}{x}}{\frac{3}{x}}=\frac{2}{x}*\frac{x}{3}=\frac{2}{3}\]
I divided the numerator by x as well: I did x/x = 1 in the numerator, and divided the whole denominator by x
if you divide the numerator, you MUST also divide the denominator!
awesome, this is just amazing. never would have figured out this.
you will learn a neat theorem called "L'hopital's rule" later which will make the computation even easier :)
k. thanks for your time. im now 'a fan' haha.
Hehe awesome ;) Good luck with everything

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