dgamma3
limit question. lim (x - > 5) 1/(x-5)^4
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dgamma3
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\[\lim_{x \rightarrow 5} 1/(1+5)^4\]
dgamma3
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woops. the 1 is an x
kirbykirby
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It would be infinity
kirbykirby
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oh
dgamma3
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why so?
kirbykirby
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Do you know L'Hopital's rule
dgamma3
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nope. is it the only way?
dgamma3
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the answer is infinity. just have no clue how to work it out.
kirbykirby
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Ok what you can do is:
kirbykirby
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wait i just noticed
kirbykirby
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u mean it's x-5 in the denominator right? otherwise it wouldn't be infinity
dgamma3
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yes, x-5
kirbykirby
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\[\lim_{x \rightarrow 5} \frac{x}{(x-5)^4}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x-5)^4}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x-5)^2(x-5)^2}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x-5)(x-5)(x-5)(x-5)}\]\[=\lim_{x \rightarrow 5} \frac{x/x}{(x/x-5/x)(x/x-5/x)(x/x-5/x)(x/x-5/x)}\]\[=\lim_{x \rightarrow 5} \frac{1}{(1-5/x)(1-5/x)(1-5/x)(1-5)}\]
kirbykirby
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oops the last factor should also be (1-5/x)
dgamma3
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awesome, thanks for your effort! just wondering line 6. why are you diving everything by x?
kirbykirby
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I hope you see that the bottom factors are (1-1) when substituting x=5, so you get like "1/0" -> infinity
kirbykirby
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hm it's true that was a useless step
kirbykirby
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I was thinking that it was going to x->infinity at first, but i realized it was x->5 after lol
dgamma3
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ahhh, yes. i see that. thank you. is that a legal move though? because you are changing the function?
dgamma3
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for example, (x-5) doesn't equal (1-5/x)
dgamma3
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sorry, i guess it is. just would have never thought of that!
kirbykirby
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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
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I divided the numerator by x as well: I did x/x = 1 in the numerator, and divided the whole denominator by x
kirbykirby
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if you divide the numerator, you MUST also divide the denominator!
dgamma3
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awesome, this is just amazing. never would have figured out this.
kirbykirby
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you will learn a neat theorem called "L'hopital's rule" later which will make the computation even easier :)
dgamma3
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k. thanks for your time. im now 'a fan' haha.
kirbykirby
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Hehe awesome ;) Good luck with everything
dgamma3
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cheers