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kubapl

Calculus Help!

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Hello, I'm taking a university level calculus course this semester. I haven't done math since collage 3 years ago. Anyways we are currently solving limit problems and I'm able to figure out a bunch in my homework but I'm stuck on these.
post-139265-0-63493600-1294601764_thumb.png
1) lim 1/x - 1/2 / x-2 x->2
2) lim (4+h)^3 -64 / h h->0

I guess my big thing is I'm not sure what to do in situations when I have fractions or when dividing with just 1 variable.

Any help would be great!

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Damn, can't link the wiki Article due to special Characters.
Anyways, search for L'Hôpitals Rule on google

wink.gif

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Actually, you don't need[font="Arial"] l'Hôpital's rule[/font]. Just simplify the expressions in order to remove the denominator.

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Actually, you don't need[font="Arial"] l'Hôpital's rule[/font]. Just simplify the expressions in order to remove the denominator.


[font=verdana, arial, helvetica, sans-serif][size=2]Yeah we haven't covered L'Hopital's rule yet. I did attempt to simplify and cancel out both of these without any luck. My answers don't match with whats in the back of the textbook.[/font]

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L'Hopital's rule is soooo convenient isn't it? However, there's a problem.

I assume that he/she is taking Calculus 1, and limits are done a week or two before derivatives. If he/she does not know how to take a derivative of a function, he/she certainly cannot do L'Hopital's rule :P.

Also, most teacher's don't really like L'Hopital's rule to be used (except when they're teaching it) when other methods are possible because it hides what's really going on.

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Anyway, here's what I got for your problems.

1. Simplify (1/x - 1/x)/(x - 2) to (2-x)/(2x^2 - 4x). Next, factor a -1 out of the numerator and a 2x out of the denominator resulting in [-(x - 2)]/[2x(x - 2)]. This simplifies to -1/2x. The limit of -1/2x as x approaches 2 is -1/4.

2. First, multiply out (4 + h)^3. This results in h^3 + 12h^2 + 48h + 64. Which, in the whole problem results in (h^3 + 12h^2 + 48h + 64 - 64)/h. This of course simplifies to h^2 + 12h + 48. The limit of h^2 + 12h + 48 as h approaches 0 is 48.

For those that are curious, these answers check out using L'Hopital's rule. :P

Kubapl, if you have any questions, feel free to ask.

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@ Ziel
NO!! To the mathematical question in your sig:

Extracting the parts that matter:
1. men give 30
2. clerk gets backs 5 out of 30 to return...so clerk has 5...........men can be concluded as having 25 out of thirty now
3. clerk pockets 2, gives men 3. men now have 25 + 3 = 28.
4. Total 28 + 2. Not a very challenging question.

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@ Ziel
NO!! To the mathematical question in your sig:

Extracting the parts that matter:
1. men give 30
2. clerk gets backs 5 out of 30 to return...so clerk has 5...........men can be concluded as having 25 out of thirty now
3. clerk pockets 2, gives men 3. men now have 25 + 3 = 28.
4. Total 28 + 2. Not a very challenging question.


It's not meant to be mathematically true, it's meant to be a fun brain teaser :P.

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2. First, multiply out (4 + h)^3. This results in h^3 + 12h^2 + 48h + 64. Which, in the whole problem results in (h^3 + 12h^2 + 48h + 64 - 64)/h. This of course simplifies to h^2 + 12h + 48. The limit of h^2 + 12h + 48 as h approaches 0 is 48.


Another way to do it, would be to directly factorize (4+h)³ - 64.
We know that a³ - b³ = ( a - b )( a² + ab + b² ).
Identifying a :=( 4 + h ) and b := 4, we can write:
( 4 + h )³ - 64 = [ ( 4 + h ) - 4 ] [ ( 4 + h )² + ( 4 + h )4 + 16 ]
which is equal to:
h[ ( 4 + h )² + ( 4 + h )4 + 16 ]

The h's are simplified and we're done with the limit.

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