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dgamma3
 3 years ago
limit question. lim (x  > 5) 1/(x5)^4
dgamma3
 3 years ago
limit question. lim (x  > 5) 1/(x5)^4

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dgamma3
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 5} 1/(1+5)^4\]

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1It would be infinity

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1Do you know L'Hopital's rule

dgamma3
 3 years ago
Best ResponseYou've already chosen the best response.0nope. is it the only way?

dgamma3
 3 years ago
Best ResponseYou've already chosen the best response.0the answer is infinity. just have no clue how to work it out.

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1Ok what you can do is:

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1u mean it's x5 in the denominator right? otherwise it wouldn't be infinity

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow 5} \frac{x}{(x5)^4}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x5)^4}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x5)^2(x5)^2}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x5)(x5)(x5)(x5)}\]\[=\lim_{x \rightarrow 5} \frac{x/x}{(x/x5/x)(x/x5/x)(x/x5/x)(x/x5/x)}\]\[=\lim_{x \rightarrow 5} \frac{1}{(15/x)(15/x)(15/x)(15)}\]

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1oops the last factor should also be (15/x)

dgamma3
 3 years ago
Best ResponseYou've already chosen the best response.0awesome, thanks for your effort! just wondering line 6. why are you diving everything by x?

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1I hope you see that the bottom factors are (11) when substituting x=5, so you get like "1/0" > infinity

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1hm it's true that was a useless step

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1I was thinking that it was going to x>infinity at first, but i realized it was x>5 after lol

dgamma3
 3 years ago
Best ResponseYou've already chosen the best response.0ahhh, yes. i see that. thank you. is that a legal move though? because you are changing the function?

dgamma3
 3 years ago
Best ResponseYou've already chosen the best response.0for example, (x5) doesn't equal (15/x)

dgamma3
 3 years ago
Best ResponseYou've already chosen the best response.0sorry, i guess it is. just would have never thought of that!

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1Nope :) 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
 3 years ago
Best ResponseYou've already chosen the best response.1I divided the numerator by x as well: I did x/x = 1 in the numerator, and divided the whole denominator by x

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1if you divide the numerator, you MUST also divide the denominator!

dgamma3
 3 years ago
Best ResponseYou've already chosen the best response.0awesome, this is just amazing. never would have figured out this.

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1you will learn a neat theorem called "L'hopital's rule" later which will make the computation even easier :)

dgamma3
 3 years ago
Best ResponseYou've already chosen the best response.0k. thanks for your time. im now 'a fan' haha.

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1Hehe awesome ;) Good luck with everything
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