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

dgamma3 Group TitleBest ResponseYou've already chosen the best response.0
woops. the 1 is an x
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
It would be infinity
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
Do you know L'Hopital's rule
 one year ago

dgamma3 Group TitleBest ResponseYou've already chosen the best response.0
nope. is it the only way?
 one year ago

dgamma3 Group TitleBest ResponseYou've already chosen the best response.0
the answer is infinity. just have no clue how to work it out.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
Ok what you can do is:
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
wait i just noticed
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
u mean it's x5 in the denominator right? otherwise it wouldn't be infinity
 one year ago

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

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
oops the last factor should also be (15/x)
 one year ago

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

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

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
hm it's true that was a useless step
 one year ago

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

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

dgamma3 Group TitleBest ResponseYou've already chosen the best response.0
for example, (x5) doesn't equal (15/x)
 one year ago

dgamma3 Group TitleBest ResponseYou've already chosen the best response.0
sorry, i guess it is. just would have never thought of that!
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
Nope :) If you divide the top and bottom by x, it is perfectly valid (it's like you are dividing by "1") Example: \[\frac{2}{3} = \frac{\frac{2}{x}}{\frac{3}{x}}=\frac{2}{x}*\frac{x}{3}=\frac{2}{3}\]
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
I divided the numerator by x as well: I did x/x = 1 in the numerator, and divided the whole denominator by x
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
if you divide the numerator, you MUST also divide the denominator!
 one year ago

dgamma3 Group TitleBest ResponseYou've already chosen the best response.0
awesome, this is just amazing. never would have figured out this.
 one year ago

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

dgamma3 Group TitleBest ResponseYou've already chosen the best response.0
k. thanks for your time. im now 'a fan' haha.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
Hehe awesome ;) Good luck with everything
 one year ago
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