Evaluate the limit as h -> 0 of:
[tan (pi/6 + h) - tan(pi/6)]/h
I thought the answer was √3/3, or tan(pi/6, but apparently that is wrong, any tips here?
12 year old thread but still saved my grade Thanks
the basic definition of the derivative of any function f(x) is
Lim [f(x+h) - f(x)]/h as h --> 0
this is exactly the pattern we are looking at.
So what they are asking for is the derivative of
tan x when x = pi/6
let y = tan x
then dy/dx = sec^2 x
= sec^2 (pi/6) or sec^2 30º
= 4/3
( cos 30º = √3/2
sec 30º = 2/√3
sec ^2 30º = 4/3 )
Thanks, I just forgot to do the dy/dx of tan x.
did not save my grade this is unreadable
To evaluate this limit, we can make use of the tangent addition formula:
tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)*tan(B))
In this case, we have A = π/6 and B = h. Let's rewrite the expression using the tangent addition formula:
[tan (π/6 + h) - tan(π/6)]/h = [(tan(π/6) + tan(h))/(1 - tan(π/6)*tan(h)) - tan(π/6)]/h
Now, let's simplify this expression further:
= [tan(π/6) + tan(h) - tan(π/6)*(1 - tan(π/6)*tan(h))]/h
= [tan(π/6) + tan(h) - tan(π/6) + tan(π/6)*tan(π/6)*tan(h)]/h
= [tan(h) + tan(π/6)*tan(π/6)*tan(h)]/h
= [tan(h) + tan^2(π/6)*tan(h)]/h
Next, we can substitute the value of tan(π/6):
= [tan(h) + (√3/3)^2*tan(h)]/h
= [tan(h) + (1/3)*tan(h)]/h
= [(1 + 1/3)tan(h)]/h
= (4/3)*tan(h)/h
Now, as h approaches 0, tan(h)/h approaches 1. This can be proven using the limit definition of the derivative of tan(x). Therefore, we can conclude that:
lim(h->0) (4/3)*tan(h)/h = (4/3)*(lim(h->0) tan(h)/h) = (4/3)*1 = 4/3
Hence, the correct answer to the limit is 4/3, not √3/3 (which is tan(π/6)).