Solve for K for Continuity of f(2) at @=0

[Hygelac]

Hi, I have just one more problem :) here it is:

(because I don't know how to do theta, "@" will equal theta)

Find a value of K so that f(2) is continuous at @=0

( != means "not equal")

f(@) = ( (2sin@)/@ , @ !=0 )
       ( 5k , @=0 )

f(@) is a piecewise function

I don't know how to get started on this problem, any help would be much appreciated :)

 

[vsage]

The simplest way would be to set 5k = lim(@->0)[2sin(@)/@]

Using l'hopital's rule, lim(@->0)[2sin(@)/@] = lim(@->0)[2cos(@)/1]

I think you can do the rest :)

 

[Hygelac]

Whops, it shoudl be f(@) not f(2)

So,

5k = (2sin(@))/@

k = [(2sin(@))/@]/5

Would that work, or did I totally miss your point?

 

[vsage]

Whops, it shoudl be f(@) not f(2)

So,

5k = (2sin(@))/@

k = [(2sin(@))/@]/5

Would that work, or did I totally miss your point?

I think you may have because that doesn't work. that function is not defined at @ = 0 so you have to use limits. k = 1/5 * limit[@->0](2sin(@)/@)