Solving the Limit of Tangent Function at h=0

[chukie]

Solve:
lim x->0 (tan 3(x+h)-tan(3x))/h

i hv no clue where to start =(

 

[rootX]

err.. tan (3h)/h?

simply plugging in 0 for x..

edit: are you sure x is approaching 0?

initially, I thought it's h..

then the answer would have been 3+3tan(3x)^2
and you had to do some mess with identites..
http://www.clarku.edu/~djoyce/trig/identities.html

 

[Dick]

Did you mean lim h->0??

 

[chukie]

Did you mean lim h->0??

sry, yes i mean lim h->0

 

[Dick]

Look up a formula for tan(A+B) and apply it to tan(3x+3h). Then simplify, take your limit and use some trig.

 

[rootX]

It is actually very simple.. don't even need to any trig after simplifying tan(3x+3h) ..

should factor out things.. and they would cancel out nicely.

And, one more thing tan(x)/x = 1 .. (which is simple to prove is you know sin(x)/x =1 as x-->0)

 

[Dick]

It is actually very simple.. don't even need to any trig after simplifying tan(3x+3h) ..

should factor out things.. and they would cancel out nicely.

And, one more thing tan(x)/x = 1 .. (which is simple to prove is you know sin(x)/x =1 as x-->0)

Yep. You don't have to use any trig. But using sec^2(A)=1+tan^2(A) would put it in the simpler form listed in books.

 

[chukie]

kk thanks i got it =)