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!
Calculus Help!
Author
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.
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!
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!
Damn, can't link the wiki Article due to special Characters.
Anyways, search for L'Hôpitals Rule on google
Anyways, search for L'Hôpitals Rule on google
Actually, you don't need[font="Arial"] l'Hôpital's rule. Just simplify the expressions in order to remove the denominator.
Author
Actually, you don't need[font="Arial"] l'Hôpital's rule. 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.
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
.
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.
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
.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.
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.
Kubapl, if you have any questions, feel free to ask.
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.
Kubapl, if you have any questions, feel free to ask.
@ 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.
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.
@ 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
.
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.
This topic is closed to new replies.
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