simple math question

[madeupfacts]

what is the limit of this?
sin(Kx)/sin(x) as x appraoch zero

 

[minendo]

42

 

[KLin]

Originally posted by: minendo
42

/thread

 

[RichUK]

 

[flyboy84]

K or 0, not sure

 

[Legendary]

http://en.wikipedia.org/wiki/L%27Hospital%27s_rule

 

[Ika]

Wow, you're taking calculus?

 

[yosuke188]

If it's a simple math question, why ask us?

 

[madeupfacts]

Originally posted by: Legendary
http://en.wikipedia.org/wiki/L%27Hospital%27s_rule

tx

 

[blueshoe]

Answer is K

 

[DrPizza]

Originally posted by: madeupfacts

Originally posted by: Legendary
http://en.wikipedia.org/wiki/L%27Hospital%27s_rule

tx

Weird... I'd have found that problem to be too easy to expect someone to come to this forum for help on - IF they had covered L'Hopital's rule already... Given the time of the year, I'm guessing that madeupfacts is in the middle of the chapter on limits that precedes the chapter that finally gets to the definition of the limit and all the juicy mathematics.

Anyway, if that's the case, they you've probably proven in class that the limit of sinx/x as x approaches 0 is 1 by using the squeezing theorem.
The homework problems typically have things like limit of sin5B/B as B approaches zero. The "steps" to find this are
multiply top and bottom by 5:

lim 5sin5B/5B

Then let x = 5B
as B approaches 0, x approaches 0

by substitution, it becomes lim 5sinx/x which is 5 * 1 = 5.
(the number of steps that are written down varies greatly depending on the prof)

Now, in your problem: sinkx/sinx
(we know that the limit is as x approaches 0)

Multiply the top and bottom by 1/x

It becomes (sinkx/x) / (sinx/x)
Do what I did above; multiply the numerator by k/k and it becomes

ksinkx/kx over sinx/x
The limit of sinkx/kx is 1, the limit of sinx/x is one, and you're left with k*1/1 = k



(lmao, my students will be doing these types of problems this week; except that since I realize that L'Hopital's rule is so much nicer to use, I don't emphasize these problems too much. Although, a problem like this one definitely falls within the level of difficulty that might be expected on their first test.

 

[madeupfacts]

Originally posted by: DrPizza

Originally posted by: madeupfacts

Originally posted by: Legendary
http://en.wikipedia.org/wiki/L%27Hospital%27s_rule

tx

Weird... I'd have found that problem to be too easy to expect someone to come to this forum for help on - IF they had covered L'Hopital's rule already... Given the time of the year, I'm guessing that madeupfacts is in the middle of the chapter on limits that precedes the chapter that finally gets to the definition of the limit and all the juicy mathematics.

Anyway, if that's the case, they you've probably proven in class that the limit of sinx/x as x approaches 0 is 1 by using the squeezing theorem.
The homework problems typically have things like limit of sin5B/B as B approaches zero. The "steps" to find this are
multiply top and bottom by 5:

lim 5sin5B/5B

Then let x = 5B
as B approaches 0, x approaches 0

by substitution, it becomes lim 5sinx/x which is 5 * 1 = 5.
(the number of steps that are written down varies greatly depending on the prof)

Now, in your problem: sinkx/sinx
(we know that the limit is as x approaches 0)

Multiply the top and bottom by 1/x

It becomes (sinkx/x) / (sinx/x)
Do what I did above; multiply the numerator by k/k and it becomes

ksinkx/kx over sinx/x
The limit of sinkx/kx is 1, the limit of sinx/x is one, and you're left with k*1/1 = k



(lmao, my students will be doing these types of problems this week; except that since I realize that L'Hopital's rule is so much nicer to use, I don't emphasize these problems too much. Although, a problem like this one definitely falls within the level of difficulty that might be expected on their first test.

ok tx, that helps more.
Are you a real doctor?

 

[chuckywang]

Originally posted by: madeupfacts
what is the limit of this?
sin(Kx)/sin(x) as x appraoch zero

K, use L'Hopital's rule.