prove lim 4x/x-4=-4 as x goes to 2

- anonymous

prove lim 4x/x-4=-4 as x goes to 2

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- schrodinger

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- Psymon

Just plug in 2 for x and then solve. Thats all you need to do.

- anonymous

that's the easiest way but we need to show f(x)-L

- zzr0ck3r

\[let \space \epsilon\space>0\\let\space \delta=8\epsilon\\then\\|x-2|<\delta\implies|\frac{4x}{x-4}+4|=|\frac{4x+4x-16}{x-4}|=|\frac{8(x-2)}{x-4}|=\\|\frac{|8(x-2)|}{|x-4|}|\le|8(x-2)=8|x-2|<\epsilon\\qed\]

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## More answers

- zzr0ck3r

make sense @psymon ?

- zzr0ck3r

bah sorry wrong tag, make sense @jhaero

- Psymon

I didnt know you could ignore the |x-4| part on the bottom was all.

- zzr0ck3r

its not ignoring it, we just know its a positive number so we know the number is bigger without it.

- zzr0ck3r

there is a absolute vale bar missing on the last line but I think its obvious.

- Psymon

So what is kind of the valid reason that we can just cut it down to 8|x-2| < epsilon? I mean, I know you say bigger without it, but....I guess Im not sure why that does or doesnt matter when we decided to not write it further.

- Psymon

Does it not matter at all what that |x-4| denominator would have been? Like if it were |x+8| or something else?

- zzr0ck3r

\[<=\]

- zzr0ck3r

snap I see what you are saying....

- zzr0ck3r

since we are around 2 for the limit, we know that the denom is > 1

- Psymon

Wait, it is? Despite that plugging in 2 makes the denominator negative?

- zzr0ck3r

||

- Psymon

Oh, right, after we apply the absolute value bars.

- Psymon

And because the denominator is greater than 1, the effect of the denominator can only be to make the whole entire thing smaller. Therefore if we just prove the numerator portion of 8|x-2|, we essentially prove the whole thing.

- zzr0ck3r

correct

- Psymon

Alrighty, cool, thanks. This is a bunch of old rehash stuff. Since we're on the subject, forever ago there was a problem using this proof and I never had it explained to me. I just remembered it, lol. So the problem was:
\[\lim_{x \rightarrow 2}x ^{2}=4 \]So you set it up like normal and then you factor it into |x+2||x-2| But I never knew what to do from there .

- zzr0ck3r

I think im wrong...

- Psymon

Yeah, thats why I never liked epsilon-delta, haha.

- zzr0ck3r

it works but I think we have another delta to fix the \x-4|

- zzr0ck3r

then to take the min...its been to long lol

- Psymon

Ikr? x_x

- zzr0ck3r

im hoping dape can set us straight before I go research some of my notes... but that is taking a very long time:P

- zzr0ck3r

and I got to get on a plane to Rome in 2 hours:P

- Psymon

I had a pathetic calc 1 professor, so my notes on material this far back are worthless, lol.

- dape

Loosely speaking, the general pattern in these limit-proving exercises is that you want to manipulate the inequality
\[|x-a|<\delta\]
so that the left-hand side looks like
\[|f(x)-L|\]
this will provide you with a inequality involving \(\delta\) which tells you how to pick the \(\delta\) as a function of \(\epsilon\) so the inequality always holds.
I will take the problem that @jhaero had as an example, first notice that
\[\left|\frac{4x}{x-4}+4\right|=\left|\frac{8x-16}{x-4}\right|=\frac{8|x-2|}{|x-4|}<8|x-2|\]
Now we can manipulate the \(\delta\)-inequality as such:
\[|x-2|<\delta\Rightarrow8|x-2|<8\delta\]
So we now see that if we pick \(\delta=\epsilon/8\), then
\[\epsilon>8|x-2|>\left|\frac{4x}{x-4}+4\right|.\]
So for this choice of \(\delta\), the inequalities are always satisfied and we have proved the limit.

- zzr0ck3r

ok that is the same thing I did. but does not answer the step about \[\frac{|8(x-2)|}{|x-4|}<8\x-2|\]

- zzr0ck3r

\[8|x-2|\]

- dape

\(|x-4|>1\), so if we replace the denominator by 1, we clearly get a bigger number.

- zzr0ck3r

ok I should have just left it alone:)

- zzr0ck3r

started to question myself....

- zzr0ck3r

so how is what you did different than what I did?

- Psymon

So how about the one I had? I guess I had no idea how to continue with mine:
\[\lim_{x \rightarrow 2}x ^{2}\]When you set it up and get to |x+2||x-2|, I have no idea what to do, lol.

- dape

You picked the wrong \(\delta\), it looks like you just made a mistake, but I also wanted to clarify the process of picking a suitable \(\delta\).

- zzr0ck3r

o yeah I meant 1/8 woops

- zzr0ck3r

ty @dape want to look at psy prob now?

- dape

That was sloppy of me.

- zzr0ck3r

http://www.purplemath.com/learning/viewtopic.php?f=14&t=790

- zzr0ck3r

did you just try and set up an "easy" example for yourself?

- zzr0ck3r

if so thats pretty funny, this one is tricky:)

- dape

I confused the factors.

- Psymon

My text did something kind of funky, Im not sure where it came from (because its a math text, duh) So it said: "For all x in the interval (1,3), x+2<5 and thus |x+2| < 5. So, letting delta be the minimum of epsilon/5 and 1, it follows that whenever you have 0 < |x-2| < delta, |x-2||x+2| < (epsilon/5)*5 = epsilon.
I guess I never followed what the text was trying to say O.o

- dape

Yeah, that's pretty clever. What is it that you don't understand?

- dape

If I make it more explicit in the end you maybe see it,
\[|x+2||x-2|<5|x-2|<5\delta=\epsilon\]

- Psymon

Well, I guess it came up with an interval that had 2 centered. Im assuming it did this so it could give some sort of definition to the extra |x+2|portion of the the problem....SO it replaces |x+2| with 5. Hmm....I guess I see it if I look at it long enough, lol. Id have never understood that strategy without seeing you write that sub in, thanks, lol.

- dape

You can use just about whatever tricks you can think of (that is valid) to pick a \(\delta\) that works, the hard part is coming up with a "trick" that works, once you have it it's usually really easy to see. You could also make a smaller interval in your prob, like \(|x+2|<4.5\) or whatever, we are only really interested in the region in the direct vicinity of \(a\) (so 2 in your case).

- zzr0ck3r

in those cases you might have 2 deltas, and then you take the min{of both deltas}

- zzr0ck3r

min{}, max{}are a must in many problems

- zzr0ck3r

but we are allowed to choose by definition so its all good:)

- Psymon

Yeah, I just never had a professor worth a damn for calc 1. It was a ton of self-teaching and this epsilon-delta stuff just kinda slipped through the cracks. Never was that great at it and never becamefroced toget good at it. I completely forgot about my confusion with this stuff until this question came up, so Im glad it finally makes some damn sense, haha x_x

- dape

To be honest it's pretty useless to be good at using the epsilon-delta thing for limits that are easily calculated, it's real usefulness lies in proving general or important results in analysis. So if you understand the concept you're golden.

- Psymon

Yeah, Ill be curious to see that when the time comes. Still got a ways to go in math, lol. Looking forward to it.

- anonymous

Here is a step by step solution: http://www.symbolab.com/solutions/limits?query=%5Clim_%7Bx%5Cto2%7D%5Cleft(%5Cfrac%7B4x%7D%7Bx-4%7D%5Cright)

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